What I did wrong in finding the radius of convergence for this problem?

Question

I'm asked to find the radius of convergence for summation from n = 1 to $\infty$ of $\sum n!x^n/(4n^n)$. I used the ratio test because of the factorial, and I eventually got a whole bunch of cancellations that just left me with $|x|$, and so if $|x| < 1$, $R=1$, but this answer is apparently wrong. I redid the problem a couple times very slowly, but I kept getting the same answer. Did I make a mistake somewhere, or am I using the wrong approach?

Edit: here is my approach

$$\lim\left|\frac{(n+1)n!x^nx}{4(n+1)^n(n+1)} \frac{4n^n}{n!x^n}\right|$$

The (n+1) terms cancel, the n! terms cancel, the 4's cancel, the n^n's cancel, the x^n terms cancel, leaving |x|

You made a mistake. Without seeing your approach, we can't say what mistake it was. I guess it was related to $\bigl(\frac{n+1}{n}\bigr)^n$. — Daniel Fischer, Oct 27 '16 at 14:05
The $n^n$ terms cancel? I only see an $n^n$ in the numerator and $(n+1)^n$ in the denominator. They don't cancel. — Thomas Andrews, Oct 27 '16 at 14:10
The leading power of the denominator will be an n^n because if you were to FOIL the (n+1)^n, the other terms in that FOIL expansion would be insignificant for large n's. Thus, you'd have an n^n/n^n. Why is this wrong? I now understand why @Crostul's works, but that's now how I originally thought of it. — Aleksandr Hovhannisyan, Oct 27 '16 at 14:10
That's true if the exponent is constant - $\frac{n^3}{(n+1)^3}\to 1$. But is is not true when the exponent changes. @AleksandrH — Thomas Andrews, Oct 27 '16 at 14:20

score 1 · Answer 1 · answered Oct 27 '16 at 14:44

I'm not sure what you did where you wrote "here is my approach" (though I think there are several mistakes...), but it seems to be pretty direct:

$$a_n:=\frac{n!}{4n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{4(n+1)^{n+1}}\cdot\frac{4n^n}{n!}=\frac{n^n}{(n+1)^n}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e$$

and thus the convergence radis is $\;R=e\;$ and we have convergence for

$$\frac{|x|}e<1\iff |x|<e\iff -e<x<e$$

Thomas Andrews · Answer 2 · 2016-10-27T14:58:15.050

0

It's actually not hard to show that $(n+1)^n\geq 2n^n$ for $n\geq 1$.

The first two terms of $(n+1)^k$ are $n^k + k\cdot n^{k-1}$. So $(n+1)^k\geq n^k+k\cdot n^{k-1}$ when $n\geq 0$ and $k\geq 1$.

When $k=n$, then $(n+1)^n\geq n^n+n\cdot n^{n-1}=2n^n$.

This in turn means that $\left(\frac{n}{n+1}\right)^n\leq \frac{1}{2}$ for all $n\geq 1$. In particular, the limit cannot be $1$, so you cannot "cancel" $n^n$ and $(n+1)^n$.

In fact, $$\left(\frac{n}{n+1}\right)^n\to e^{-1}.$$

edited Oct 27 '16 at 14:58

answered Oct 27 '16 at 14:28

Thomas Andrews

177,126

Oh boy, I think that stuff's a bit too advanced for me. It's easier to understand L'Hospital's Rule with a related function. Thanks anyway! – Aleksandr Hovhannisyan Oct 27 '16 at 14:38

What I did wrong in finding the radius of convergence for this problem?

2 Answers2