# Flawed reasoning somewhere when calculating the radius of convergence for a power series

I keep getting the wrong answer working the following problem out this way. I'm hoping someone could tell me where my reasoning is flawed. I'm surely missing something. The problem is to find the radius of convergence for:

\begin{align} f(x) &= \sum_{k=1}^{\infty} \frac{k!x^k}{k^k} \end{align}

To do so, I use the ratio test to determine interval of convergence:

\begin{align} R &= \lim_{k \to \infty} \left\vert \frac{(k+1)!x^{k+1}}{(k+1)^{k+1}} \cdot \frac{k^k}{k!x^k} \right\vert && \text{definition of ratio test} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)!}{k!} \cdot \frac{x^{k+1}}{x^k} \cdot \frac{k^k}{(k+1)^{k+1}} \right\vert && \text{arrange like terms} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)}{} \frac{ k!}{k!} \cdot \frac{x^{k+1}}{x^k} \cdot \frac{k^k}{(k+1)^{k+1}} \right\vert && \text{definition of factorial} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)}{}\cdot \frac{x^{k+1}}{x^k} \cdot \frac{k^k}{(k+1)^{k+1}} \right\vert && \text{simplify} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)}{} \cdot \frac{x}{}\frac{x^{k}}{x^k} \cdot \frac{k^k}{(k+1)^{k+1}} \right\vert && \text{properties of exponentiation} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)}{} \cdot \frac{x}{} \cdot \frac{k^k}{(k+1)^{k+1}} \right\vert && \text{simplify} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)}{} \cdot \frac{x}{} \cdot \frac{}{(k+1)} \frac{k^k}{(k+1)^{k}} \right\vert && \text{properties of exponentiation} \\\\ &= \lim_{k \to \infty} \left\vert \frac{(k+1)}{(k+1)} \cdot \frac{x}{} \cdot \frac{k^k}{(k+1)^{k}} \right\vert && \text{arrange like terms} \\\\ &= \lim_{k \to \infty} \left\vert \frac{x}{} \cdot \frac{k^k}{(k+1)^{k}} \right\vert && \text{simplify} \\\\ &= \lim_{k \to \infty} \left\vert \frac{x}{} \cdot \frac{k^k}{k^k + c_1k^{k-1} + c_2k^{k-1} + ... + c_{k-1}k+1} \right\vert && \text{binomial expansion} \\\\ &= \lim_{k \to \infty} \left\vert \frac{x}{} \cdot \frac{1}{1 + \frac{c_1}{k} + \frac{c_2}{k^2} + ... + \frac{c_{k-1}}{k^{k-1}} + \frac{1}{k^k}} \right\vert && \text{divide num and denom by } k^k \\\\ &= \lim_{k \to \infty} \left\vert \frac{x}{} \cdot \frac{1}{1} \right\vert && \text{let k go to } \infty \\\\ \lim_{k \to \infty} \left\vert x \right\vert &= \left\vert x \right\vert \\\\ R &= \left\vert x \right\vert \end{align}

However... computers and the textbook tell me this limit evaluates to $$R=\frac{x}{e}$$

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented Jul 16 at 5:30

Notice where you have:

$$\lim_{k \rightarrow \infty} \Big(\frac{k}{k+1}\Big)^k = \frac{1}{1}$$

This is not correct!

This limit is $$1/e$$, which will lead to the answer that your textbook provides.

I have linked you to WolframAlpha, in which this computation is performed; to prove that this is correct may involve how you have defined $$e$$. For example, some define the reciprocal of this limit to be the definition of $$e$$. In other words, some define:

$$\lim_{k \rightarrow \infty} \Big(\frac{k+1}{k}\Big)^k = \lim_{k \rightarrow \infty} \Big(1 + \frac{1}{k}\Big)^k = e$$

One might similarly suspect that the limit in parentheses goes to $$1$$; after all, the $$1/k$$ term appears to be going to $$0$$. But, this is not the case for the entire expression: In the race between $$1/k \rightarrow 0$$ and the exponent $$k \rightarrow \infty$$, the term in question converges to $$e = 2.718281828459045 \ldots$$.

• [downvote? idgi! please leave a comment if something is off, thanks] Commented Jul 18 at 0:10

I'd like to explain where you went wrong. As has been explained by Benjamin Dickman, the limit

$$\lim_{k\to\infty}\left(\frac{k}{k+1}\right)^k$$

is the original culprit, but what you did with it isn't exactly wrong, but the notation got the better of you and hid a crucial detail.

Doing the binimial expansion step is "half right and half wrong", when you substituted $$(k+1)^k$$ with $$k^k + c_1k^{k-1}+ c_2k^{k-2} + \ldots+ c_{k-1}k+1.$$

The problem here is twofold:

1. Those $$c_i$$ aren't constant, they depend on $$k$$. It's easy to see that $$c_1= {k\choose 1} = k, c_2={k\choose 2}=\frac{k(k-1)}2$$, for example, and this goes on.
2. Even the number of those coefficients is increasing with $$k$$, you have $$k-1$$ of them.

Usually when you do some binomial expansions, the exponent ($$k$$ in our case) is not part of the binom to expand. Usually the exponent is a fixed value, and in that case there are always the same number of coefficients and they are the same, so your notation with $$c_i$$ makes sense, the $$c_i$$ are actually constants.

In that case you can do the routine of "divide num and denom by $$k^k$$" and evaluate that numerator and denominator separately tend to some value.

But that does not work when the $$c_i$$ are not constant and depend on the variable you take the limit over!

Another problem is that the evaluation of the limit for the denominator also does not work as you did when the number of summands is variable and not bounded.

As a simple example

$$\lim_{k\to\infty} 1 = 1$$

is trivial. But let me replace the $$1$$ with a sum of $$k$$ values $$\frac1k$$:

$$\lim_{k\to\infty} \sum_{i=1}^k \frac1k = 1$$

On the left hand side you have a sum of values (all $$\frac1k$$), each such value undoubtedly going to $$0$$ when $$k\to\infty$$. But the sum of them does not go to $$0$$. That's because while each summand itself is small, you have many of them.

If you have a sum with a variable number of summands, you cannot get the limit of the sum by taking the limit of each term and sum them. That's because you have (in the limit) infinitely many terms.

• Thank you for the thorough explanation! I had my suspicions that it was to do with trying to expand the denominator like that, but could not figure out why. The (hidden by me) interdependency between the exponent and the binomial coefficients makes perfect sense. The division by $k^k$ step doesn't actually result in terms that reduce to zero. Commented Jul 16 at 22:46
• @Sean: that's right. Actually if you try to figure out what happens to those terms when you divide by $k^k$ you find that they approach $1 + \frac{1}{2!} + \frac{1}{3!} + \dots$ which is one way to see that the limit is $\frac{1}{e}$. Commented Jul 16 at 23:08

Your sin was to write a sum with indeterminate coefficients, and not care about what happens when $$k$$ tends to infinity.

Better evaluation of these coefficients would have shown you that the terms do not vanish.

(I am unfortunately unable to add a comment)

Apart from the issue of evaluating the limit the radius of convergence must not depend on x. From you result you can conclude that the series converges for x/e < 1, that is x < e = R.

• Pleas don't post comments as answers. Commented Jul 17 at 9:57
• Once you get enough reputation, you'll be able to leave comments. The radius of convergence doesn't depend on $x$, but the result of the ratio test can. In this case, OP thought the ratio test resulted in $|x|$, so that the radius of convergence would have been 1. You are correct that the ratio test should have resulted in $|x|/e$, so that the radius of convergence is $e$. Commented Jul 17 at 21:35