using CalculusWithJuliaSquared
plotly()17 Sequences and series
This section uses the following add-on package:
CalculusWithJuliaSquared re-exports Plots and Symbolics, so neither needs a using of its own. Limits here are taken with symlim, which returns the value and the route it took — so every answer says how it was reached.
This section expands on limits of infinite sequences and their sums.
17.1 Definitions
An infinite sequence is simply an infinite, ordered set of terms \(a_1, a_2, a_3, \dots\); the initial term may be indexed by any fixed integer, not just \(1\). We may refer to the individual terms, \(a_n\), or the whole sequence \(\{a_n\}\).
The partial sum is the sequence \(s_n = a_1 + a_2 + \cdots + a_n = \sum_{i=1}^na_i\). Again, the starting index need not be \(1\).
A series is a sum of an infinite sequence \(\sum_{i=1}^\infty a_i = a_1 + a_2 + a_3 + \cdots\).
As mentioned, a sequence converges to \(L\) if for any \(\epsilon\) we can find an \(M\) such that if \(n > M\) then \(|a_n - L| < \epsilon\). A non-convergent sequence is called divergent. Series too may be convergent of divergent, with details to come.
17.2 Examples
Some examples of sequences are:
- A constant sequence \(a_n = c\) will converge to \(c\).
- The sequence \(a_n = 1/n\) clearly converges to \(0\).
- Similarly, \(a_n = n^k\) diverges if \(k \geq 0\) and converges (to \(0\)) when \(k < 0\).
- The sequence \((1 + 1/n)^n\) has a limit of \(e\), which comes up in compound interest calculations and numerous other places. This limit can be used to define the value of \(e\).
- The sequence \(a_n = r^n\) will converge to \(0\) if \(|r| < 1\); converge to \(1\) when \(r=1\) and diverge otherwise.
- The sequence \(a_n = \cos(n \pi)\), \(n\geq 1\) diverges. It is said to oscillate. (It is \(-1,1,-1,1,-1,\dots\).)
17.2.1 Some limit theorems for sequences
The limit theorems apply to limits of sequences as well:
Other limit theorems
We mention a few other limit theorems for sequences.
One fact about convergent series is
Convergent series are bounded.
(That is, there exists an \(M>0\) with \(-M \leq a_n \leq M\) for all \(n\).)
Not all bounded sequences converge; however:
The sequence \(a_n = (1 + 1/n)^n\) is both monotone increasing and bounded, hence convergent. As mentioned, it converges to \(e\). That this limit has a known value is not a result of the theorem, which only says some value exists.
A subsequence of an infinite sequence is an infinite sequence chosen by taking only some of the terms (along the same order). A simple example would be \(b_n = a_{2n}\), which would take every other term of the sequence \(\{a_n\}\). More generally if \(\phi(n)\) is an increasing function with integer values, then \(b_n = a_{\phi(n)}\) would be a formal way to define a subsequence.
A sketch is to consider the interval \(I_1 = [-M,M]\). It has infinitely many values of the sequence in it when \(M\) is a bound. Divide this interval in half. There is a choice of \(I_2\) so that it also contains infinitely many points of the sequence. (Maybe both do, in which case either choice can be made.) This splitting and choosing can be repeated to create an infinite sequence of nested intervals \(\{I_n\}\) each containing infinitely many points of \(\{a_n\}\) and each of length \(M/2^{n-2}\). As these intervals are nested the left-hand endpoints are bounded and increasing, hence convergent.
There may be many different convergent subsequences, this just identified one.
Finally, this following fact can be used to reverse the pedagogical approach of defining limits by starting with limits restricted to sequences.
17.3 Series
As defined, a series is an infinite sum of a sequence: \(a_1 + a_2 + \cdots\). The partial sums of a sequence are \(s_n = a_1 + a_2 + \cdots + a_n\). The series is convergent (divergent) if the sequence of partial sums, \(\{ s_n \}\), is convergent (divergent).
Example: the geometric series
A ubiquitous series is the geometric series for a value of \(r\). This is the sum \(s = \sum_{i=0}^\infty r^i\). The partial sums satisfy
\[ s_n = \sum_{i=0}^n r^i = 1 + r + r^2 + \cdots + r^n = \frac{1 - r^{n+1}}{1 - r}, \]
a formula that comes from multiplying out the sum by \(1-r\).
If \(|r| < 1\) then \(s\) is convergent and \(s_n \rightarrow (1 - r)^{-1}\). If \(|r| \geq 1\) the series is divergent.
Example: geometric series adjacent
Consider now a related sum \(s = \sum_{i=0}^\infty i \cdot r^i = \sum_{i=1}^\infty i \cdot r^i\), assuming \(|r| < 1\). What does this converge to?
The partial sums also have a (lesser) known formula:
\[ \sum_{i=1}^n i\cdot r^i = r \frac{1 - (n+1)r^n + nr^{n+1}}{(1 - r)^2}. \]
Rather than take this on trust, we can derive it — and the derivation is short, because it is the geometric sum with one derivative applied.
Symbolics has no symbolic summation: there is no way to write \(\sum_{i=1}^{n}\) with a symbolic upper bound, so the sum cannot simply be handed over and evaluated. But it does not need to be. Start from the geometric sum of the previous section,
\[ \sum_{i=0}^{n} r^i = \frac{1 - r^{n+1}}{1 - r}, \]
and differentiate both sides with respect to \(r\). On the left, \(\frac{d}{dr} r^i = i\,r^{i-1}\) term by term — the sum is finite, so this is just linearity, with none of the convergence care an infinite sum would demand. Multiplying by \(r\) turns \(i\,r^{i-1}\) into \(i\,r^i\), which is the sum we want:
\[ \sum_{i=1}^{n} i\,r^{i} = r\cdot\frac{d}{dr}\left[\frac{1 - r^{n+1}}{1-r}\right]. \]
Differentiation is untroubled by the symbolic exponent — a restriction that bites when taking limits, not derivatives:
@variables r n
Symbolics.derivative(r^(n+1), r)
\[ r^{n} ~ \left( 1 + n \right) \]
So the whole claim can be checked. Take the derivative, multiply by \(r\), subtract the proposed formula, and the difference should be zero:
@variables r n
geo = (1 - r^(n+1)) / (1 - r)
proposed = r * (1 - (n+1)*r^n + n*r^(n+1)) / (1-r)^2
Symbolics.simplify_fractions(r * Symbolics.derivative(geo, r) - proposed)
\[ \frac{ - r^{1 + n} ~ r - r^{1 + n} ~ n ~ r + r^{2} ~ r^{n} + r^{2} ~ r^{n} ~ n}{1 - 2 ~ r + r^{2}} \]
That does not look like zero — but read the numerator. It is
\[ -r^{1+n}\cdot r \;-\; n\,r^{1+n}\cdot r \;+\; r^{n}\cdot r^{2} \;+\; n\,r^{n}\cdot r^{2}, \]
and \(r^{1+n}\cdot r\) and \(r^{n}\cdot r^{2}\) are both \(r^{n+2}\), so the first term cancels the third and the second cancels the fourth. The expression is zero. Symbolics will not see it, because combining those two requires adding exponents when one of them is symbolic, and it does not do that.
This is worth knowing in its own right, because it will happen again: when an expression that should collapse refuses to, look for a repeated power with a symbolic exponent. The fix is to stop asking the system to reason about \(r^{n}\) at all — give that whole quantity a name, and what is left is ordinary algebra in \(r\) and \(R\):
@variables r n R
geo = (1 - r^(n+1)) / (1 - r)
proposed = r * (1 - (n+1)*r^n + n*r^(n+1)) / (1-r)^2
d = r * Symbolics.derivative(geo, r) - proposed
simplify(substitute(d, Dict(r^n => R, r^(n+1) => R*r)))
\[ 0 \]
Exactly 0, and for every \(n\) — not for a handful of sampled values. The substitution \(R = r^{n}\), \(r^{n+1} = R\,r\) carries no assumption; it just says the same thing in symbols the simplifier can finish.
The excluded case is \(r = 1\), where the formula’s \((1-r)^2\) denominator vanishes and the sum is just \(1 + 2 + \cdots + n\). Since we are assuming \(|r| < 1\) that is not our case.
As \(n \rightarrow \infty\) for \(|r| < 1\) this expression goes to \(s = r / (1-r)^2\). It would diverge otherwise.
Example: Sum of inverse factorials
Consider these two sequences:
\[ s_n = \sum_{k=0}^n \frac{1}{k!}, \quad p_n = \left(1 + \frac{1}{n}\right)^n \]
We know \(p_n \rightarrow e\). We will see that it also follows that \(s_n \rightarrow e\). That is the series for the sequence \(a_k = 1/k!\) converges to \(e\).
First we see that \(e\) is in the ballpark.
We bound each term of \(s_n\) for \(k \geq 2\)
\[ a_k = \frac{1}{k!} \leq \frac{1}{(k(k-1))} = \frac{1}{k-1} - \frac{1}{k}, \]
With this, we can identify a telescoping sum:
\[ \begin{align*} s_n &= \sum_{k=0}^n a_k\\ &= a_0 + a_1 + \sum_{k=2}^n a_k\\ &\leq 2 + \sum_{k=2}^n \left(\frac{1}{k-1} - \frac{1}{k}\right) \\ &= 2 + \left(\frac{1}{1} - \frac{1}{n}\right) = 3 - \frac{1}{n} < 3. \end{align*} \]
This is in agreement with \(s = e\). To get that value takes more effort and different bounds.
The binomial theorem applied to \(p_n\) gives:
\[ \begin{align*} p_n &= \left(1 + \frac{1}{n}\right)^n\\ &= \sum_{k=0}^n {n \choose k} \frac{1}{n^k} 1^{n-k}\\ &= \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k}\\ &= \sum_{k=0}^n \frac{1}{k!}\frac{n!}{(n-k)!}\frac{1}{n^k}\\ &= \sum_{k=0}^n \frac{1}{k!} \cdot \left(1 - \frac{1}{n}\right) \cdot \left(1-\frac{2}{n}\right) \cdot \cdots \cdot \left(1 - \frac{k-1}{n}\right)\\ &= \sum_{k=0}^n \frac{1}{k!} b_{n,k} \end{align*} \]
We see that \(p_n \leq s_n\) by noting \(b_{n,k} \leq 1\) as each term in it is.
Next, following a reference, we establish the bound \(s_n - 3/(2n) \leq p_n\).
We have by multiplying out:
\[ \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) = 1 - \frac{1 + 2}{n} + \frac{1\cdot 2}{n} > 1 - \frac{1 + 2}{n}. \]
Assume this pattern holds for some \(k\), then we have the induction step
\[ \begin{align*} \left(1 - \frac{1}{n}\right) &\left(1 - \frac{2}{n}\right)\cdot\left(1 - \frac{k}{n}\right)\cdot\left(1 - \frac{k+1}{n}\right)\\ &> \left(1 - \frac{1 + 2 + \cdots + k}{n}\right)\left(1 + \frac{k+1}{n}\right)\\ &= 1 - \frac{1 + 2 + \cdots + k + (k+1)}{n} + \frac{(1+2+\cdots+k)(k+1)}{n}\\ &> 1 - \frac{1 + 2 + \cdots + k + (k+1)}{n} \end{align*} \]
So the inequality holds for any \(k\) by induction. In our bound for \(b_{n,k}\) we have this sum \(1 + 2 + \cdots + (k-1) = (k-1)k/2\). Together this gives:
\[ \begin{align*} p_n &= \sum_{k=0}^n \frac{1}{k!}b_{n,l}\\ &> \sum_{k=0}^n \frac{1}{k!} \cdot \left(1 - \frac{(k-1)k}{2n}\right)\\ &= s_n - \sum_{k=0}^n \frac{1}{k!}\frac{(k-1)k}{2n}\\ &= s_n - \frac{1}{2n} \sum_{k=2}^n \frac{1}{(k-2)!}\\ &= s_n - \frac{1}{2n} s_{n-2}\\ &> s_n - \frac{3}{2n} \end{align*} \]
The last inequality using the earlier bound on \(s_n\).
Together these observations give the bounds:
\[ s_n - 3/(2n) \leq p_n \leq s_n \leq p_n + 3/(2n). \]
If we know \(p_n \rightarrow e\), then by the squeeze theorem, we get \(s_n \rightarrow e\).
17.3.1 Non-negative terms
The examples above required a bit of problem-specific work to get a value for a series.
Sometimes the value isn’t important—just the question of convergence is. There are some general things that are the case for series to understand convergence.
First we consider only sequences with non-negative terms.
This says if \(a_n\) does not converge to \(0\) then \(s_n\) diverges. It is definitely not the case that a sequence that converges to \(0\) will lend itself to a convergent series. A famous example would be \(\sum_{i=1}^n 1/i\) which diverges. The partial sums of this series are termed the harmonic series and have the property that \(s_n = \ln(n) + \gamma + 1/(2n) + \epsilon_n\) where \(e_n \rightarrow 0\) and \(\gamma \approx 0.5772\) is a constant termed the Euler-Mascheroni constant. (See MathConstants.γ.)
Fix any \(N > 0\), the partial sums with \(n>N\) satisfy:
\[ s_n = \sum_{k=1}^n a_k =\sum_{k=1}^N a_k + \sum_{k=N}^n a_k = s_N + (s_n - s_N) \]
The limit as \(n \rightarrow \infty\) does not depend on the constant \(s_N\).
This can be used to prove, for example, that if a series based on a non-negative sequence converges, any series based on a subsequence will also converge.
Example
Does the following sequence of partial sums converge of diverge:
\[ s_n = \sum_{i=1}^n \frac{i}{i^2 - i^{-3}}? \]
Diverge. We can see that the \(i\)th term is bounded below:
\[ \frac{i}{i^2 - i^{-3}} = \frac{1}{i^1 - i^{-4}} > \frac{1}{i} \]
As each term is bounded below by a sequence whose series diverges, the partial sums in question must diverge. This follows as \(i^{-4} > 0\) for any positive \(i\) so \(i - i^{-4} < i\); the reciprocal reverses the bound.
Example
Does the following sequence of partial sums converge or diverge?
\[ s_n = \sum_{i=1}^n \sqrt{i} \cdot r^i, \quad 0 < r < 1? \]
As \(\sqrt{i} \leq i\) for \(i\geq 1\), so \(\sqrt{i} \cdot r^i \leq i r^i\). We can use the comparison test to say \(s_n\) converges, as we earlier saw the bounding series does.
There are other tests that, when applicable, are more direct and avoid needing to identify a bound.
For the first case, identify an \(r\) with \(|L| < r < 1\). Then for some \(M > 0\) and any \(n > M\) it follows that \(a_{n+1} \leq a_n r\). For any \(n>0\) the convergence only depends on values after \(n\). As such, we shift things to assume for all \(n\) that \(a_{n+1} \leq a_n r\). Iterating, we get
\[ a_{n+1} \leq a_n r \leq a_{n-1}r^2 \leq \cdots \leq a_1 r^n \]
By the comparison test, the series \(\sum a_k\) converges, since \(0 < r < 1\).
The case for \(L > 1\) is similar, only we find a lower bound on each term.
For the case \(L=1\)—which is where most hard problems fall—we have examples where the series converges or diverges. The harmonic series is a divergent example, the series of the sequence \(\{1/n^2\}\) is convergent (with limit \(\pi^2/6\)), though we haven’t proved that (cf. Basel problem).
A similar type of theorem involves powers of the terms in the sequence
The proof is similar. Consider the first case with \(L < r < 1\). Then \((a_n)^{1/n} < r\) implies \(a_n < r^n\) so the partial sums (possibly after ignoring initial terms and some shifting) satisfy \(a_0 + a_1 + a_2 + \cdots + a_n < r^0 + r^1 + r^2 + \cdots + r^n\). The geometric series converges under the assumptions. By the comparison test so must the series in question.
The same two examples (\(a_n = 1/n\) and \(a_n = 1/n^2\)) give examples of divergent and convergent series when \(L = 1\).
This test is a consequence of a more general integral test which will be discussed later, below we offer a specific proof.
When \(p = 1\) this clearly diverges, it being the harmonic series.
When \(p < 1\), we have \(i^p < i\) so \(1/i^p > 1/i\). By the comparison test (to the harmonic series) the series \(s\) will diverge.
To prove this series converges when \(p > 1\), we split the sum up into different pieces from \(2^k\) to \(2^{k+1}-1\). Call this sum \(t_k\) then \(s = \sum_k t_k\) with
\[ t_k = \sum_{i=2^k}^{2^{k+1}-1} \frac{1}{n^p} \leq 2^k \cdot \frac{1}{(2^{k})^p} \leq \left(\frac{2}{2^p}\right)^k \]
The above replaces each term with the smallest value over \(2^k\) to \(2^{k+1}-1\) and then multiplies by the number of terms.
For \(p > 1\), the value of \(2/2^p < 1\). That is, \(s\) is bounded by a geometric series hence convergent.
Example
Consider this series
\[ \sum_{n=1}^\infty \frac{n^k}{5^n}, \] for some positive integer \(k\).
Will this converge?
The ratio of \(a_{n+1}\) to \(a_n\) is:
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^k}{5^{n+1}} \frac{5^n}{n^k} = \left(\frac{n+1}{n}\right)^k \frac{1}{5} \rightarrow \frac{1}{5} \]
As the limit of this ratio is less than \(1\), the series converges.
Example
Let \(a\) be a positive number and consider this series
\[ \sum_{n=1}^\infty \frac{a^n}{n^n} \]
Does this converge?
The root test is easy to apply given the two powers of \(n\):
\[ (a_n)^{1/n} = \left(\frac{a^n}{n^n}\right)^{1/n} = \frac{a}{n} \rightarrow 0 \]
That this is less than \(1\) says the series converges.
Example
Consider this series:
\[ \sum_{n=3}^\infty \frac{1}{n^{3/2} \log(n)} \]
This isn’t exactly in the format for the \(p\)-series test, but we note that \(\log(n) \geq 1\) for \(n\geq 3\). So
\[ a_n = \frac{1}{n^{3/2} \log(n)} \leq \frac{1}{n^{3/2}} \]
The series \(\sum n^{-3/2}\) converges by the \(p\)-series test, hence the series in question must also converge.
17.3.2 General series
If there is no assumption that \(a_n\) is non-negative, then some different behaviours are possible.
First, we say that a series is absolutely convergent if \(\sum |a_n|\) converges.
A series which is convergent but not absolutely convergent is termed conditionally convergent.
Example
Is this series convergent?
\[ s = \sum_{k=1}^\infty \frac{\sin(n)}{n^4} \]
We show that the series is absolutely convergent by the comparison theorem.
We have
\[ |a_n| = \frac{|\sin(n)|}{n^4} \leq \frac{1}{n^4} \]
Since \(p=4 > 1\) the series is absolutely convergent by the \(p\)-series test:
However, not all convergent series are also absolutely convergent. A key example is the alternating harmonic series:
\[ s = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \]
This series is conditionally convergent—not absolutely convergent.
Why follows immediately from a more general statement.
To visually see this, consider Figure 17.1.
Example: reordering
Take the sequence \(\{ a_n \}\). This is simply a list of values with some indicated order. Let a reordering be any sequence formed by traversing the values in a different order. There are two key facts:
If the series based on the sequence is absolutely convergent, then a series based on any reordering will be absolutely convergent.
If the series is conditionally convergent but not absolutely convergent then a reordering may converge (conditionally) to a different value or even diverge.
The case of the latter is the alternating harmonic series. to see it diverges, we note that the subsequences of both negative and positive terms diverge. Call these \(\{b_n\}\) and \(\{c_n\}\). To see that we can make these diverge, we note that for any \(N>0\) and \(L\) we can find \(k\) so that \(b_N + b_{N+1} + \cdots + b_{N+k} > L\). Now for each \(i\), we take a consecutive subsequence of values of \(b_n\) which sum to \(i + |c_i|\). Then the subsequence formed by a group of \(b\)s followed by \(c_i\) will have partial sums always bigger than \(i\), hence divergent.
17.4 Power series
A polynomial of degree \(n\) is defined by its coefficients through: \(p(x) = a_0 + a_1\cdot x + a_2 \cdot x^2 + \cdots + a_n \cdot x^n\) where \(a_n\) is non zero. The polynomial as a function of \(x\) may be shifted through \(p(x-c) = a_0 + a_1\cdot (x-c) + a_2 \cdot (x-c)^2 + \cdots + a_n \cdot (x-c)^n\).
Pushing this, let \(\{a_n\}\) be a sequence. We define a power series around \(c\) by:
\[ \sum_{n=0}^\infty a_n (x - c)^n. \]
A typical case is \(c=0\). For any fixed \(x\) this is simply a series. Convergence is seen to depend on the value of \(x\).
Suppose the series \(\sum a_n\) is absolutely convergent. Then there are some values around \(c\) for which the series converges (\(x=c\) is one). The radius of convergence is a value \(r\) for which
- if \(|x-c| < r\) then the power series converges absolutely; and
- if \(|x-c| > r\) the power series diverges.
The root test indicates why such a value exists.
Suppose we consider the term \(b_n = a_n (x-c)^n\). Then
\[ (|b_n|)^{1/n} = (|a_n|)^{1/n} |x-c| \]
If \((|a_n|^{1/n}) \rightarrow L\) then the root test indicates if the power series converges absolutely or not. In particular. If \(L|x-c| < 1\) it converges and if \(L|x-c| > 1\) it diverges. That is if \(|x-c| < 1/L\) or \(|x-c| > 1/L\). Taking \(r=1/L\) yields the radius of convergence. (For cases where the limit does not exist, a relaxed version of the root test with the limit inferior is applicable.)
The ratio test can also be used to establish the radius of convergence provided:
\[ \frac{|a_{n+1}||(x-c)^{n+1}|}{|a_n||(x-c)^n|} = \frac{|a_{n+1}|}{|a_n|} |x-c| \rightarrow L |x-c|. \]
Again, \(r=1/L\).
Examples
Consider the power series
\[ 1 + \frac{x}{1} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
This has \(c=0\) and \(a_n = 1/n!\). The ratio test simplifies to
\[ \frac{|x-c|^{n+1}}{(n+1)!} \cdot \frac{n!}{|x-c|^n} = \frac{|x-c|}{n} \rightarrow 0 \]
Hence \(r=\infty\) and the power series is always absolutely convergent.
Consider the power series
\[ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \]
This has term \(a_n = (-1)^{n} x^{2n+1}/(2n+1)\), \(n\geq 0\). The root test for absolute convergence has
\[ (|a_n|)^{1/n} = \frac{x^{2 + 1/n}}{(2n+1)^{1/n}} \rightarrow x^2 \]
Provided \(|x| < 1\) the power series will converge.
Both these examples are related to Taylor series for some function.
17.5 Questions
Question
Which of these sequences converge? (The options define the \(i\)th element in the sequence.)
Question
Which of these sequences diverge?
Question
Which of these series converge?
Question
Which of these series diverge?
Question
Some terms grow faster than others. For example, it is well known that exponential growth is greater than polynomial growth. Here we look at a few forms.
A polynomial term is like \(n^k\) for some power \(k>0\). We know from discussion on horizontal asymptotes that \(n^k\) grows faster than \(n^j\) if \(k > j\).
What about comparing a polynomial term to an exponential term? Suppose \(r > 1\); we compare \(a_n = r^n\) against \(b_n = n^k\).
Written as it stands, \(r^n/n^k\) has the variable in both exponents, and a symbolic exponent is the one thing that blocks every exact route — the order of \(n^k\) is \(k\) itself, so there is nothing to rank until \(k\) is known. Rewriting into log-exponential form moves the symbols out of the exponents and into coefficients, which is exactly the form the limit engine is built for:
\[ \frac{r^n}{n^k} = e^{\,n\log r \;-\; k\log n}. \]
Fixing \(r = 2\) as a concrete instance of \(r > 1\), but leaving \(k\) symbolic:
@variables n k
symlim(exp(n*log(2) - k*log(n)), n, Inf)(Inf, :gruntz)That this is \(\infty\) for every \(k\) at once indicates that the exponential term \(r^n\) grows faster than any “polynomial” term \(n^k\).
Compare the exponential term \(r^n\) to a factorial term \(n!\). (There is no symbolic factorial to hand this to, so compare the ratio of successive terms, as the ratio test does: \(\frac{r^{n+1}}{(n+1)!}\cdot\frac{n!}{r^n} = \frac{r}{n+1}\), which goes to \(0\).)
What do you conclude?
Compare the growth of a term like \(n^n\) to \(n!\) — again through successive terms, which gives \(\left(1 + \tfrac{1}{n}\right)^{n}\), a limit met earlier in this section. What do you conclude?
Question
Consider the sequence defined by
\[ a_n = \frac{n!}{n^n}, \quad n \geq 0. \]
Symbolics has no factorial of a symbolic argument — factorial(n) errors outright, and routing through \(\Gamma(n+1)\) does not help, since \(\Gamma(n+2)/\Gamma(n+1)\) will not cancel either. But the ratio test does not need one: the factorials cancel by hand, which is how the test is done on paper anyway.
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \frac{(n+1)\,n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^{n}. \]
The single cancellation \((n+1)!/n! = n+1\) does all the work. What is left is a limit symlim can take:
@variables n
symlim((n/(n+1))^n, n, Inf)(0.36787944117144233, :composition)The route is :composition, and it names the argument. This is \(f^g\) with \(f \to 1\) and \(g \to \infty\) — the indeterminate form \(1^\infty\) — so it is rewritten as \(e^{\,n\log(n/(n+1))}\), the inner limit is found to be \(-1\), and the answer is \(e^{-1}\) because \(e^x\) is continuous there. That last step is the Composition rule stated among the limit theorems at the top of this section: \(\lim f(a_n) = f(\lim a_n)\). Seeing the route named is the point — it is the difference between being handed a number and being shown which theorem produced it.
Based on this output, does the sequence converge?
Question
Compute
\[ \sum_{n=0}^\infty \frac{2}{3^n} + \frac{4}{5^n} \]
Question
Compute
\[ \sum_{n=2}^\infty \left(\frac{1}{2}\right)^n \]
Question
Consider the series
\[ \sum_{n=1}^\infty \frac{n}{e^n} \]
This series
Question
Consider the series
\[ \sum_{n=1}^\infty (-1)^n \cdot \frac{n}{2n - 3} \]
This series
Question
Consider the power series about \(c=0\) formed from the sequence
\[ a_n = (-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!} \]
That is, \(s = x - x^3/3! + x^5/5! - \cdots\). Find the radius of convergence.
As above, cancel the factorials by hand first. With \(a_n = x^{2n+1}/(2n+1)!\),
\[ \frac{a_{n+1}}{a_n} = \frac{x^{2n+3}}{(2n+3)!}\cdot\frac{(2n+1)!}{x^{2n+1}} = \frac{x^{2}}{(2n+2)(2n+3)}, \]
since \((2n+3)! = (2n+3)(2n+2)\,(2n+1)!\). Then:
@variables x n
symlim(x^2/((2n+2)*(2n+3)), n, Inf)(0, :reciprocal)The limit is \(0\) whatever \(x\) is — the denominator grows without bound while the numerator does not depend on \(n\) — so the ratio test gives convergence for every \(x\).
The radius of convergence is:
Question
Consider the series based on the terms
\[ a_n = (-1)^{n}\frac{1}{n}(x-1)^n \]
Find the radius of convergence.
Here the powers cancel rather than the factorials, but the move is the same one — the symbolic exponent has to go before a limit can be taken:
\[ \frac{a_{n+1}}{a_n} = \frac{(x-1)^{n+1}}{n+1}\cdot\frac{n}{(x-1)^{n}} = (x-1)\cdot\frac{n}{n+1}. \]
@variables x n
symlim((x-1) * n/(n+1), n, Inf)(-1 + x, :gruntz)
The limit is \(x-1\). The ratio test asks about \(|a_{n+1}/a_n|\), so it is \(|x-1|\) that must be less than \(1\) — the series converges when \(|x - 1| < 1\), giving a radius of \(1\) about \(c=1\).
The radius of convergence is:
Question
Consider the series
\[ \sum_{k=1}^\infty k\cdot x^k = x + 2x^2 + 3x^3 + \cdots \]
We can show directly, without needing a formula for the partial sums as before, that this series converges whenever \(|x| < 1\).
We want to know how \(r^n\) compares with \(n\) when \(r = 1 + \delta\) for some \(\delta > 0\). Rewriting \(r^n\) as \(e^{\,n\log r}\) as before, and asking with \(\delta\) left symbolic:
@variables n delta
symlim(exp(n*log(1 + delta)) - n, n, Inf)(nothing, :parameter_dependent)symlim refuses, and the refusal is the lesson. Symbolics has no assumptions system: there is no way to say \(\delta > 0\), so nothing in the expression rules out \(\delta < 0\) — and for a negative \(\delta\) we would have \(r < 1\), making \(r^n\) vanish while \(-n\) runs to \(-\infty\). The limit is \(+\infty\) on one branch and \(-\infty\) on the other, so there is no single answer to give, and :parameter_dependent says exactly that.
Pinning \(\delta\) to a positive value asks the question we actually meant:
@variables n
symlim(exp(n*log(3//2)) - n, n, Inf) # δ = 1/2, so r = 3/2 > 1(Inf, :gruntz)and to a negative one confirms the other branch is real, not hypothetical:
@variables n
symlim(exp(n*log(1//2)) - n, n, Inf) # r = 1/2 < 1(-Inf, :gruntz)
WarningA returned answer can be right for a question you did not ask
Nothing about \(\log(1+\delta)\) says which sign it has, so an engine willing to answer here would answer for one branch and say nothing about which. That is more dangerous than a refusal: -Inf looks like an ordinary result, and it is the exact opposite of the claim this exercise is building toward.
The general habit, whenever a limit carries a free parameter: pin it to a representative value and compare against the free-symbol answer. If they differ, the parameter’s range is part of the problem and has to be stated. SymPy states it through positive=True; here it is stated by choosing a value and saying so in the text.
Since \(\log r > 0\) whenever \(r > 1\), the same computation runs for any such \(r\): \(n\log r\) outgrows \(\log n\), so \(e^{n \log r}\) outgrows \(n\).
Suppose \(r^n > n\), then \(nx^n \leq n|x|^n < r^n |x|^n < (r|x|)^n\). If \(|x| < 1\) why can we find \(r>1\) with \(r|x| < 1\)?
If \(r\) is so chosen, then by the comparison theorem the sum is eventually bounded by a geometric series.
Question
Let \(s(x) = \sum_{i=1}^\infty i^2 x^i\) be a power series. The ratio test for convergence considers:
\[ |\frac{a_{n_+1}}{a_n}| = \frac{(n+1)^2 x^{n+1}}{n^2x^n} = \frac{(n+1)^2}{n^2} |x|. \]
This has a limit as \(n \rightarrow \infty\). Use this limit to find the radius of convergence. What is the value?
Now for fixed integer \(k > 2\) let \(s(x) = \sum_{i=1}^\infty i^k x^i\) be a power series. The ratio test again can be used to find the radius of convergence. What does it find?
Question
Absolutely convergent series have some properties
If \(\sum a_n\) and \(\sum b_n\) are absolute convergent then \(\sum (a_n + b_n)\) will be absolutely convergent.
if \(\sum a_n\) and \(\sum b_n\) are absolute convergent then \(\sum (a_n \cdot b_n)\) will be absolutely convergent.
To see the first we could use the triangle inequality:
\[ |a_n + b_n| \leq |a_n| + |b_n| \]
So that the partial sum
\[ \sum_{k=1}^n |a_n + b_n| \leq \sum_{k=1}^n |a_n| + \sum_{k=1}^n |b_n| \]
Why do each of these sums converge?
For the second one, if \(\sum b_n\) is absolutely convergent, then we can assume \(|b_n| < 1\) eventually. Hence, eventually
\[ |a_n b_n| \leq |a_n| \cdot 1 \]
Since \(\sum a_n\) is absolutely convergent by the comparison theorem so will be \(\sum a_n b_n\).
Why is \(|b_n| < 1\) eventually?
Question
The Cauchy criteria for convergence of a series \(s = \sum a_k\) is that \(s\) converges if for any \(\epsilon > 0\) there exists an \(N\) such that if \(n,m \geq N\) then \(|s_n - s_m| < \epsilon\) and vice versa. This basically says the terms at the end eventually contribute a negligible amount to the sum when convergent.
Why might this be useful?
Question
The following extends the comparison test:
Basically the limit says both sequences \(\{a_n\}\) and \(\{b_n\}\) go to \(0\) at the same rate so their series are somewhat comparable.
To prove this, fix an \(\epsilon\) so \(c-\epsilon > 0\). Then eventually
\[ \left| \frac{a_n}{b_n} - c \right| < \epsilon \]
Rearranging gives \(a_n < (c+\epsilon) b_n\).
If \(\sum a_n\) diverges why must \(\sum b_n\) diverge?
Another rearrangement gives \(b_n < a_n / (c - \epsilon)\). If \(\sum a_n\) is convergent why must \(\sum b_n\) be convergent?
Question
Let \(s = \sum_{n=0}^\infty a_n \cdot (x-c)^n\) and \(t = \sum_{n=0}^\infty b_n \cdot (x-c)^n\).
Guess what this power series is:
\[ \sum_{n=0}^\infty (a_n + b_n) \cdot (x-c)^n \]
Guess what this power series is:
\[ \sum_{n=0}^\infty \left(\sum_{i=0}^n (a_i \cdot b_{n-i})\right) \cdot (x-c)^n \]
For completeness, Wikipedia gives this formula for division \(s/t = \sum_{n=0}^\infty d_n \cdot (x-c)^n\) where
\[ d_0 = \frac{a_0}{b_0} \]
and \[ d_n = \frac{1}{b_0^{n+1}} \cdot \begin{vmatrix} a_n & b_1 & b_2 & \cdots & b_n\\ a_{n-1} & b_0 & b_1 & \cdots & b_{n-1}\\ a_{n-2} & 0 & b_0 & \cdots & b_{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\\ a_0 & 0 & 0 & \cdots & b_0 \end{vmatrix} \]
The last operation is called the determinant and will be discussed later on.