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Find the interval of convergence of the following series $$\sum_{n=1}^{\infty}\frac{n^n x^n}{n!}.$$

I've found the out that the series converges absolutely for $|x|<1/e$ but diverges for $|x|>1/e.$ So, when $x=1/e,$ we have

$$\sum_{n=1}^{\infty}\left(\frac{n}{e }\right)^n\frac{1}{ n!}.$$ Also, when $x=-1/e,$ we have

$$\sum_{n=1}^{\infty}(-1)^n \left(\frac{n}{e }\right)^n\frac{1}{ n!}.$$

My question is: how do I proceed from here?

Omojola Micheal
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4 Answers4

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HINT

By Stirling approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

we have that

$$\left(\frac{n}{e }\right)^n\frac{1}{ n!}\sim \frac1{\sqrt{2 \pi n}}$$

user
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Note that $n\mapsto a_{n}\equiv(n/e)^{n}/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $\lim_{n\rightarrow\infty}a_{n}=0$. Therefore, by the alternating series test, $\sum(-1)^{n}a_{n}$ converges.

parsiad
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Stirling's approximation says that$${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}$$therefore the first sum diverges also the 2nd sum converges since $$\text{2nd sum}\sim-1+\sum_{n=1}^{\infty}\dfrac{(-1)^n}{\sqrt n}{\\=-1+\sum_{k=1}^{\infty}\dfrac{1}{\sqrt {2k}}-\dfrac{1}{\sqrt {2k+1}}\\=-1+\sum_{k=1}^{\infty}\dfrac{\sqrt {2k+1}-\sqrt {2k}}{\sqrt {2k}\sqrt {2k+1}}\\=-1+\sum_{k=1}^{\infty}\dfrac{1}{\sqrt {2k}\sqrt {2k+1}}\dfrac{1}{\sqrt {2k+1}+\sqrt {2k}}\\<-1+\sum_{k=1}^{\infty}\dfrac{1}{2k\sqrt{2k+1}}<\infty}$$which is obviously convergent

Mostafa Ayaz
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  • I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy. – Jakobian Jul 14 '18 at 20:50
  • Why then?...... – Mostafa Ayaz Jul 14 '18 at 20:51
  • Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't – Jakobian Jul 14 '18 at 20:57
  • I separated the alternating coeffs pairwise such as following$$-1+\dfrac{1}{\sqrt 2}-\dfrac{1}{\sqrt 3}+\dfrac{1}{\sqrt 4}-\dfrac{1}{\sqrt 5}+\dfrac{1}{\sqrt 6}-\dfrac{1}{\sqrt 7}+\cdots=-1+(\dfrac{1}{\sqrt 2}-\dfrac{1}{\sqrt 3})+(\dfrac{1}{\sqrt 4}-\dfrac{1}{\sqrt 5})+(\dfrac{1}{\sqrt 6}-\dfrac{1}{\sqrt 7})+\cdots$$ – Mostafa Ayaz Jul 14 '18 at 21:00
  • @Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point. – Omojola Micheal Jul 14 '18 at 21:07
  • There is a theorem stating that if $a_n>0$ , $\sum a_n$ is convergent and $b_n=a_n+o(a_n)$ then $\sum b_n$ is convergent either – Mostafa Ayaz Jul 14 '18 at 21:21
  • The first $<$ is obviously wrong. – Szeto Jul 14 '18 at 22:48
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The radius of convergence of $\sum_{n=0}^\infty a_nx^n$ is given by $R = \lim_{n\to\infty} \left|\frac{a_{n}}{a_{n+1}}\right|$, if this limit exists.

We have

$$\left|\frac{\frac{n^n}{n!}}{\frac{(n+1)^{n+1}}{(n+1)!}}\right| = \frac{n^n}{(n+1)^n} = \frac1{\left(1+\frac1n\right)^n} \xrightarrow{n\to\infty} \frac1e$$

so $R = \frac1e$, as you predicted.

mechanodroid
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