# Why does a Power series converge at its centre even if the ratio test yields a limit that does not exist?

Consider the following Power series, $${\sum}_{k=0}^{\infty}k!x^{k}$$ The ratio test yields the following result,

$$\underset{k\rightarrow\infty}{\lim}\left|\frac{\left(k+1\right)!\cdot x^{k+1}}{k!x^{k}}\right|=\underset{k\rightarrow\infty}{\lim}\left|\left(k+1\right)\cdot x\right|=x\cdot\underset{k\rightarrow\infty}{\lim}\left|k+1\right|$$

The resulting limit does not exist. In fact for $$x=0$$, it seems to yield the indeterminate form $$0\cdot\infty$$. Yet my textbook concludes that the series must converge for $$x=0$$, why is this true?

• When $x=0$, $$\sum_{k=1}^{\infty} k! x^k = \sum_{k=1}^{\infty} k! 0^k = \dotsb?$$ Apr 8 '20 at 20:29

### Key Idea

The goal is to determine the behaviour of the series $$\sum_{j=1}^{\infty} b_j |x-a|^{j}.$$ Start with elementary tools (such as direct computation), and only apply deeper results when the elementary tools fail. In the case of a power series, we can check that $$\sum_{j=1}^{\infty} b_n |x-a|^n \bigg|_{x=a} = \sum_{j=1}^{\infty} b_n 0^n = 0,$$ hence there is no reason to apply the ratio (or any other) test. Indeed, the ratio test fails to apply in this case, as it involves a division by zero.

### Further Discussion

First and foremost, recall the definition for series convergence:

Definition: Let $$\{a_n\}$$ be a sequence of real (or complex) numbers, and for each $$n \in \mathbb{N}$$, define $$S_n := \sum_{j=1}^{n} a_j.$$ If $$\lim_{n\to \infty} S_n$$ exists, then we say that the infinite series $$\sum_{j = 1}^{\infty} a_j$$ converges, and has the value $$\sum_{j = 1}^{\infty} a_j = \lim_{n\to\infty} S_n.$$

When working with power series, where the general term is of the form $$a_n = b_n (x-a)^n$$, we are often interested in determining the values of $$x$$ such that the series converges, in the sense defined above. It is "obvious" that if $$x = a$$, the series converges, as $$\sum_{j=1}^{\infty} b_j (x-a)^j = \sum_{j=1}^{\infty} b_j 0^j = \sum_{j=1}^{\infty} 0 = 0.$$ Determining other values of $$x$$ such that the series converges is a little more difficult. While it might be possible to evaluate the sum directly, the typical approach is to apply theorems from analysis to simplify things. For example, the ratio test:

Theorem: Let $$\{a_n\}$$ be sequence of real (or complex) numbers, and define $$L = \left| \frac{a_{n+1}}{a_n} \right|.$$ Assume that "the series" refers to $$\sum_{j=1}^{\infty} a_j$$. Then

1. if $$L < 1$$, the series converges absolutely,
2. if $$L > 1$$, the series diverges, and
3. if $$L = 1$$, the behaviour of the series cannot be determined by this test.

The ratio test is a tool for simplifying the process of determining whether or not a series diverges. Note that there exist cases such that the ratio test is indeterminate, and that the ratio test implicitly requires that $$a_n \ne 0$$ for all sufficiently large $$n$$ (otherwise, $$L$$ would not exist). Also note that this is a result about series in general—the application to power series is just a special case.

To apply the result to a power series, suppose that $$a_n = b_n (x-a)^n$$. Then, assuming that the limit even exists, $$L = \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to \infty} \left| \frac{b_{n+1} |x-a|^{n+1}}{b_{n}|x-a|^n} \right| = |x-a| \lim_{n\to \infty} \left| \frac{b_{n+1}}{b_n} \right|.$$ Then, by the ratio test, the power series conveges whenever $$L < 1$$, which is equivalent to $$|x-a| < \frac{1}{L} = r \qquad(\text{the radius of convergence}),$$ which gives a technique for determining the radius of convergence. Other tests of convergence will work better in other circumstances (for example, the root test is commonly used in this situation).

One final thought: in the question, it is noted that if one attempts to apply the ratio test when $$x=a$$, it is possible to arrive at an indeterminate form, e.g. when $$a_n = n! x^n$$: $$L = \lim_{n\to\infty} \left| \frac{(n+1)!x^{n+1}}{n!x^n} \right| = |x| \lim_{n\to\infty} (n+1).$$ When $$x=0$$, this is of the form $$L "=" 0\cdot \infty$$, which is an indeterminate form. Whenever we arrive at an indeterminate form, all we have learned is that the tool we are using is the wrong tool for the job. In elementary calculus, this often means applying L'Hospital's rule or computing a limit using some other method. In this case, it is an indication that the ratio test was the wrong tool in the case $$x=0$$ (note that it works perfectly well for all other $$x$$, as $$|x| \lim_{n\to\infty}(n+1) = +\infty$$).

Your ratio test is invalid at $$x=0$$ since the denominator is $$0$$. Instead note that $$\sum_{k=1}^\infty k! 0^k =0$$

• Very interesting, I didn't think of that, so I should qualify my ratio test with the condition that $x\neq0$? Apr 8 '20 at 20:30
• Yes. In order to even talk about ratio test, you need $a_n x^n \neq 0$ for all $n$ sufficiently large. In particular, $x\neq 0$ Apr 8 '20 at 20:33