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10. Convergence of Series

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One of the hallmarks of calculus is the theory of infinite series, particularly that of Taylor series and power series. This is a framework in which certain “nice” functions are expressed as $$f(x) = \sum _{n=0}^{\infty} a_n x^n$$ for a sequence of real numbers $a_n$, for instance. From a practical standpoint, this is particularly useful when computing values of certain functions, and many computer math libraries and calculators do this under the hood. Power series can, in certain scenarios, be much easier to manipulate than arbitrary functions, aiding both theory and applications.

In these scenarios, it’s important for people to know (1) for which values of $x$ does the series converge to an actual number, and (2) how many terms from the infinite series need to be added up in order to get “close enough” to the actual value. Our focus at the tail end of this class will be the first question — how can you tell if a series converges or not?

If $\sum _{n=0}^{\infty} a_n$ converges, then necessarily $\lim _{n\to\infty} a_n = 0$. If one wants to add up infinitely many numbers, those numbers must be getting smaller; this is called the divergence test. However, there are still sequences with $\lim _{n\to\infty}a_n = 0$ for which the series still diverges — the most famous example is the harmonic series $\sum _{n=1}^{\infty} \frac{1}{n} = \infty$. We have to use more sophisticated reasoning in these cases.

Most of the time, it’ll be with a comparison test — there are a small handful of “baseline” series that have been known to converge or diverge since antiquity.

• Geometric series $\sum _{n=0}^{\infty} r^n$ converge to $\frac{1}{1-r}$ whenever $\left\lvert r \right\rvert < 1$, and it diverges if $\left\lvert r \right\rvert \geq 1$.
• $p$-series of the form $\sum _{n=1}^{\infty} \frac{1}{n^p}$ converge if $p > 1$ and diverge if $p \leq 1$.

These are facts that should be burned into your mind. I think the geometric series makes sense — at the very least, if $\left\lvert r \right\rvert \geq 1$, then you’re adding infinitely many things that aren’t getting any smaller, and there’s no way that should be a finite quantity. The $p$-series should be reminiscent of the improper integrals $\int _{1}^{\infty} \frac{1}{x^p} dx$, which converge with exactly the same conditions on $p$. In fact, we’re doing an “improper Riemann sum”.

Not every series is of these forms, but you should think of using these as “references”. Does my series $\sum _{n=0}^{\infty} a_n$ resemble a geometric series in a quantitative way? Does it resemble a $p$-series in a quantitative way? This is what convergence tests do.

Our first pair of convergence tests assess how similar a given series is to a geometric series.

If the ratio of consecutive terms is $L$, then eventually the terms start looking like part of a geometric series. Each term is (more or less) a factor of $L$ away from the previous. Thus, the cases $L> 1$ and $L< 1$ reflect exactly the convergence and divergence of geometric series. Do note that when $L = 1$, it’s impossible to tell if a series converges or diverges using this test alone; you’ll need to use one of the other tests to find out.

The ratio test is particularly useful when there are powers of a constant and/or factorials involved, as these cancel nicely in the ratios.

Solution

Let $a_n = \frac{5^n}{n!}$. Then, $$\lim _{n\to\infty} \left\lvert \frac{a _{n+1}}{ a_n } \right\rvert= \lim _{n\to\infty} \frac{5 ^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n} = \lim _{n\to\infty} \frac{5}{n+1}=0.$$ Note here that $(n+1)! = 1\cdot 2 \cdots n \cdot (n+1) = n! \cdot (n+1)$.

Since $0 < 1$, the series converges by the ratio test.

Solution

Again, we’ll use the ratio test. Let $a_n = \frac{n!}{n^n}$. We have $$L = \lim _{n\to\infty} \left\lvert \frac{a _{n+1}}{a_n} \right\rvert = \lim _{n\to\infty} \frac{(n+1)!}{(n+1) ^{n+1}} \cdot \frac{n^n}{n!}.$$ Since $(n+1)! = (n+1) \cdot n!$, just like the last problem, the $n!$’s cancel each other out. The remaining $n+1$ cancels out with one power of $\left( n+1 \right) ^{n+1}$, leaving just $(n+1)^n$. Thus, we get $$L = \lim _{n\to\infty} \frac{n^n}{(n+1)^n} = \lim _{n\to\infty} \left( \frac{n}{n+1} \right)^n.$$ We have to use L’Hôpital’s rule to evaluate this limit! (Take natural logs of everything.) You’ll eventually end up with $$L = \frac{1}{e} < 1,$$ hence this series again converges by the ratio test.

There’s another test that works on a similar principle to the ratio test, but makes the connection to geometric series in a slightly different way.

Just like the ratio test, the root test uses the heuristic that $a_n \approx L^n$ when $n$ is very, very large. Thus, the conditions for convergence and divergence when $L \neq 1$ again mimick those of normal geometric series, though again one can’t use this test when $L = 1$. This test is particularly useful when there are lots of things being raised to the $n$-th power, but it’s a lot harder to apply when factorials or polynomial quantities are present.

A more direct set of convergence tests are the comparison tests, and these are often important to use when the ratio test or the root test are inconclusive.

The direct comparison test says (morally) that if $\sum _{n=0}^{\infty} b_n < \infty$, then $$\left\lvert \sum _{n=0}^{\infty} a_n \right\rvert \leq \sum _{n=0}^{\infty} \left\lvert a_n \right\rvert \leq \sum _{n=0}^{\infty} b_n < \infty.$$ Thus the series for $a_n$ is finite, i.e. it converges. Likewise, if you can bound $a_n$ from below by $b_n$ and $\sum _{n=0}^{\infty} b_n$ diverges, then $\sum _{n=0}^{\infty} a_n$ is “bigger than $\infty$”, hence must also diverge. Note that the other comparisons do not work in our favour.

The limit comparison test instead says that if $a_n$ eventually looks like a constant multiple of the sequence $b_n$, then the two series must both converge or both diverge.

These tests are most useful when the ratio and root tests are inconclusive, particularly when the summand is close to a rational function of $n$. You are usually applying these tests to determine if $\sum _{n=0}^{\infty} a_n$ converges or not, where $a_n$ is given as part of a problem. You have complete freedom to choose $b_n$, and almost always you’ll want to pick a geometric series or a $p$-series, whose convergence is easy to assess.

Solution

For problems like this, I would recommend squinting your eyes and asking, when $n$ is really big, what’s the simplest approximation I can make?

When $n$ is say $10 ^{10}$, we have $n+4 \approx n$ and $3n^3+2n+1 \approx 3n^3$. Thus, I would like to say $$\sum _{n=1}^{\infty} \frac{n+4}{3n^3+2n+1} \approx \sum _{n=1}^{\infty} \frac{1}{n^2},$$ and I know this is a convergent $p$-series. To make this rigorous, I might apply the limit comparison test: if $a_n = \frac{n+4}{3n^3+2n+1}$ and $b_n = \frac{1}{n^2}$, we have $$\lim _{n\to\infty} \frac{a_n}{b_n} = \lim _{n\to\infty} \frac{n^3+4n^2}{3n^3+2n+1} = \frac{1}{3}.$$ This is a finite, nonzero quantity, and since $\sum _{n=1}^{\infty} b_n <\infty$, the series $\sum _{n=1}^{\infty} a_n$ converges too.

Solution

Although this looks more intimidating, the exact same thought process from the previous example applies. Note that we can’t immediately apply the ratio or the root test — there’s a lot of extra junk in the way that would prevent easy simplification.

Instead, when $n$ is really big, $2n+3^n\approx 3^n$ while $14\sqrt n + \ln n + 2^n \approx 2^n$. So, I would like to say that $$\sum _{n=1}^{\infty} \frac{2n+3^n}{ 14\sqrt n + \ln n + 2^n} \approx \sum _{n=1}^{\infty} \left( \frac{3}{2} \right)^n,$$ and I know the rightmost side is a diverging geometric series.

Rather than wrangle with the limit comparison test, let’s demonstrate how to use the direct comparison test. I need to make $$\frac{2n+3^n}{14\sqrt n + \ln n + 2^n} \geq C \cdot \frac{3^n}{2^n},$$ where $C$ can be any constant. Note that I used the “squint your eyes” test to determine which direction I need the inequality to go.

In the numerator, I trivially have $2n+3^n \geq 3^n$ for all $n$. In the denominator, I have that $\sqrt n \leq 2^n$ and $\ln n \leq 2^n$ for all $n$, thus $14 \sqrt n + \ln n + 2^n \leq 16 \cdot 2^n$ for all $n$. Note that the inequality is reversed! Putting the two together, we get $$\frac{2n+3^n}{14\sqrt n + \ln n + 2^n} \geq \frac{3^n}{16\cdot 2^n}.$$ Since the series $\sum _{n=0}^{\infty} \frac{3^n}{16 \cdot 2^n}$ is a diverging geometric series, we know $\sum _{n=1}^{\infty} \frac{2n+3^n}{14\sqrt n + \ln n + 2^n}$ diverges too by the comparison test.