Де збігається степеневий ряд?

In the previous topic, we looked at convergence of power series $\sum_{n=0}^\infty c_n(x-a)^n$ for various values of $x\in\mathbb{R}$, and observed that we usually get a result like the series converges for $|x-a|<R$ and diverges for $|x-a|>R$, some $R>0$, possibly $+\infty$. Now we'll study this more closely.

Theorem: Consider the power series $\sum_{n=0}^\infty c_n(x-a)^n$, and let $\frac{1}{R}=\limsup_{n\to\infty}|c_n|^{1/n}$. Then the power series

(i) converges for $|x-a|<R$

(ii) diverges for $|x-a|>R$

(If the limsup is 0, then we set $R=+\infty$, and if the limsup is $+\infty$, we set $R=0$.) $R$ is called the radius of convergence.

Ready for the proof?

Proof: We use the root test, just like we did in the previous section when we were given a specific $c_n$. However, we use a slightly more general version of the root test; instead of just a normal limit, we use a limsup. For most cases, especially in a calculus class (rather than an analysis class), the distinction doesn't really matter here, and you can replace the limsup with an ordinary limit. $$\limsup_{n\to\infty}|c_n(x-a)^n|^{1/n}=\limsup_{n\to\infty}|c_n|^{1/n}|x-a|=|x-a|\frac{1}{R}$$ This expression is $<1$ for $|x-a|<R$, and $>1$ for $|x-a|>R$, assuming $R\in(0,\infty)$. For $R=0$, then the limit is $>1$ for every value of $x\ne a$, and so the power series diverges for every $x\ne a$. For $R=\infty$, the limit is $0$ for every value of $x$, and so the power series converges for all $x\in\mathbb{R}$.

Remark: We don't know what happens at the endpoints, i.e. where $|x-a|=R$. These cases have to be settled by a different test for series, often the limit comparison test or alternating series test.

TODO: Add examples

We discussed in the previous explanation how to find the radius of convergence of a power series $\sum_{n=0}^\infty c_n(x-a)^n$. We did this by applying the Root or Ratio test, i.e. by comparing to a geometric series. Recall that $$\frac{1}{R}=\limsup_{n\to\infty}|c_n|^{1/n},$$ and $R\in[0,+\infty]$ is called the radius of convergence. The power series converges for $|x-a|<R$ and diverges for $|x-a|>R$, which we proved in the theorem in the last section via the Root Test.

Now we'll look at ways to compute $R$. The most obvious way is to just compute $\limsup_{n\to\infty}|c_n|^{1/n}$. But depending on what $c_n$ is, this might be kind of difficult. (Just like how sometimes computing the limit using the Ratio Test is easier than calculating the limit of the $n$th root in the Root Test.)

So instead, we can use the Ratio Test. If the limit $\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|$ in the ratio test exists, then it is guaranteed to be the same as $\limsup_{n\to\infty}|c_n|^{1/n}$, and so is a valid way to compute $R$.

Consider $\sum_{n=0}^\infty c_n(x-a)^n$, and let's apply the Ratio Test instead of the Root Test: $$\lim_{n\to\infty}\frac{|c_{n+1}(x-a)^{n+1}|}{|c_n(x-a)^n|}=\lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}|x-a|,$$ and so the power series converges if $|x-a|<\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|$ and diverges if $|x-a|>\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|$. This is all assuming that $\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|$ exists or is infinite.

So we can instead compute: $$\frac{1}{R}=\limsup_{n\to\infty}|c_n|^{1/n}=\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|$$ if the limit exists (or is infinite).

Ready for examples?

Example 1: Find the radius of convergence of $\sum_{n=0}^\infty n(x-1)^n$.

We can compute $\frac{1}{R}=\limsup_{n\to\infty}|n|^{1/n}=1$, so $R=1$. Alternatively, $\frac{1}{R}=\lim_{n\to\infty}\frac{n+1}{n}=1$. Note then the series converges for $|x-1|<1$, and diverges for $|x-1|>1$.

Example 2: Find the radius of convergence of $\sum_{n=0}^\infty\frac{x^{2n}}{n3^n}$.

Caution!! Our power series is not in the form $\sum_{n=0}^\infty c_n x^n$, so the formula for $R$ doesn't work immediately. We can, however, view this as a power series in $x^2=:y$ rather than in $x$, in which case we do have the correct form $\sum_{n=0}^\infty c_ny^n$. We use the Root or Ratio test to compute $$\frac{1}{R}=\limsup_{n\to\infty}\frac{1}{n^{1/n}(3^n)^{1/n}}=\frac{1}{3},$$ so $R=3$ for the series in $y$, not $x$! Thus the series converges for $|y|<3$, i.e. for $|x^2|<3$, or $|x|<\sqrt{3}$. The radius of converge for the original power series is $R=\sqrt{3}$.

Another way to compute the radius of convergence is to just apply the Root or Ratio test, ignoring the formula for $\frac{1}{R}$. This method requires nearly the same amount of work and works in general, even when we don't have exactly the correct form.

Given a power series in $x$ centered at a point $a$, what can we say about the set of inputs for which the power series converges? Since power series are modeled off of geometric series, the root test or the ratio test comes into play! It turns out that there always exists a radius of convergence $R\ge 0$, such that if $|x-a| < R$ (i.e. if $x\in (a-R,a+R)$), then the power series converges absolutely, while if $|x-a| > R$, then the power series diverges. It's not easy, in general, to tell whether a power series converges at the points $x = a\pm R$, when $|x-a|$ equals $R$ exactly.

In practice, to find the radius of convergence, you can often use the root test or the ratio test. If you want to determine the exact interval of convergence, then the endpoints may or may not take a little extra care.