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I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^{\infty} \dfrac{n!x^n}{n^n}$.

I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$:

$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|<1$$

So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$:

$$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$

So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes:

$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$

Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$:

$$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right) =\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$

Applying L'Hôpital's rule:

$$\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}} =\lim_{n\to+\infty}\frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}} =\lim_{n\to+\infty}\left(-\frac{n}{n+1}\right)=-1$$

Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that:

$$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$.

Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is:

$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$

Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}<1$, that is, $|x|<e$.

So, I know that the series is convergent for $-e < x < e$, but I have to test whether the series is convergent at $x = e$ or $x = -e$. That is, I have to test whether $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ and $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ are convergent. Since these limits don't approach zero, I know they are both divergent, but I'm not sure how to find the limits, because of the factorial function. Also, I can't use integral test here, because of the factorial. Probably I should use comparison test, but I haven't found any divergent series to which to compare it.

Any hints?

Thank you in advance.


Edit: Using the suggestion by Ragib Zaman in the answer below, since the Taylor polynomial $P_n(x)$ of $e^x$ at $a=0$ is $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}+\cdots,$$ if we substitute $n$ for $x$ we see that $e^n>\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} > 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero.

$\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term.

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1  
well the Stirling formula should give you the equivalence $(-1)^n\sqrt{2\pi n}$. –  Raymond Manzoni Aug 1 '12 at 12:07
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Stirling formula shows that the general term of these series does not converge to zero hence they both diverge. –  Did Aug 1 '12 at 12:08
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+1 This question shows research effort... –  draks ... Aug 1 '12 at 12:16
    
Of course first you should find if the term goes to zero. If it does, then you need further work. –  GEdgar Aug 1 '12 at 12:17
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Just a short comment: $$\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n=\lim_{n\to+\infty}\left(\frac{1‌​}{1+1/n}\right)^n=\lim_{n\to+\infty}\frac{1}{\left(1+1/n\right)^n}=1/e$$ –  N. S. Aug 1 '12 at 19:09

2 Answers 2

up vote 8 down vote accepted

Since $\displaystyle e^x > \frac{x^n}{n!}$ for all $x>0, n \in \mathbb{N}$, letting $x=n$ gives $\displaystyle \frac{n!e^n}{n^n}>1$ so the terms don't even approach zero.

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Thank you for this suggestion, but why does this inequality hold? I'm not sure how to prove it. –  anonymous Aug 2 '12 at 23:13
    
@anonymous Have you seen the Taylor series of $e^x$ ? –  Ragib Zaman Aug 3 '12 at 2:08
    
OK, I see it now. –  anonymous Aug 3 '12 at 11:51
    
One more question: is the following reasoning correct? $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term. –  anonymous Aug 3 '12 at 12:24
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@anonymous Your statement that $e \left( \frac{n}{n+1} \right)^n >1$ is correct, and yes it is true that if the absolute value of the terms is increasing, the series can not converge. But that is not the best way to view why the series diverges, which is simply that the terms don't approach zero. –  Ragib Zaman Aug 3 '12 at 12:50

This may be overkill but we have $$n!\approx n^{n+1/2}e^{-n}$$ by Stiriling's formula when $n$ goes very large. Thus the original sum becomes approximately $$n^{1/2}\frac{x^{n}}{e^{n}}$$ So if $x\ge e$ this series obviously will not converge. When $0<x<e$ name $\frac{x}{e}=k<0$, then we have $$\sum n^{1/2}k^{n}$$ the ratio test give $$(\frac{n+1}{n})^{1/2}k$$ which becomes less than 1 when $n$ goes very large.

The case that $x$ is negative is similar by the alternating series test.

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Why did you overlook the $\sqrt{2\pi}$ term? –  Pedro Tamaroff Nov 10 '12 at 21:53

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