The task is to test the following series for convergence/ divergence: $$\sum_{n=1}^\infty \frac{(a+nx)^n}{n!}$$

Now, I have been able to use the Ratio Test and establish that the series converges for $x<1/e$ and diverges for $x>1/e$, but testing the series at $x=1/e$ has been a little more challenging. Could someone tell me how I might get the job done?

  • $\begingroup$ I think you will need Stirling formula. $\endgroup$ – hamam_Abdallah Jan 15 '19 at 19:47

Let $x=\frac{1}{e}$. Since $$ \frac{(a+\frac{n}{e})^n/n!}{(\frac{n}{e})^n/n!}=\left(1+\frac{ae}{n}\right)^n\to e^{ae}, $$ by the comparison test, $\sum_{n=1}^\infty a_n<\infty $ if and only if $\sum_{n=1}^\infty \frac{(\frac{n}{e})^n}{n!}<\infty$. So we may assume that $a=0$. By Stirling's formula, we have $$\lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}(\frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that $$c\le \frac{\sqrt{n}(\frac{n}{e})^n}{n!}\le C, $$or equivalently $$ \frac{c}{\sqrt{n}}\le \frac{(\frac{n}{e})^n}{n!}\le \frac{C}{\sqrt{n}}. $$ Since $\sum_n \frac{1}{\sqrt{n}}=\infty$, the series diverges for $x=\frac{1}{e}$.

If $x=-\frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|\to 0$ as $n\to \infty$. And this follows immediately from Stirling's formula: $$\begin{eqnarray} \lim_{n\to\infty}|a_n|&=& \lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{n!}\\ &=&\lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{\sqrt{2\pi n}(\frac{n}{e})^n}\\ &=&\lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\left(1-\frac{ae}{n}\right)^n=0. \end{eqnarray}$$

There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have $$ \log(1+t) = t-\frac{t^2}{2}+o(t^2). $$ This implies that there exists $\delta>0$ such that $$ \exp\left(t-ct^2\right)\le 1+t\le \exp\left(t-Ct^2\right),\quad\forall 0\le t\le \delta $$ for some $c\in (\frac{1}{2},1)$ and $C\in (0,\frac{1}{2})$. Let $$a_n = \frac{(\frac{n}{e})^n}{n!}.$$ Then $$ \frac{a_{n+1}}{a_n}=\frac{1}{e}\left(1+\frac{1}{n}\right)^n, $$ and hence $$ \exp\left(-\frac{c}{n}\right)\le\frac{a_{n+1}}{a_n}\le \exp\left(-\frac{C}{n}\right) $$ for all sufficiently large $n$. This gives for all large $n$, $$ k\exp\left(-c\sum_{j=1}^{n-1}\frac{1}{j}\right)\le a_n\le K\exp\left(-C\sum_{j=1}^{n-1}\frac{1}{j}\right) $$ for some $k>0$ and $K>0$. Using the fact that $\int_j^{j+1}\frac{dt}{t}\le\frac{1}{j}=\int_{j-1}^j\frac{1}{j}dt\le\int_{j-1}^j\frac{dt}{t}$ for $j\ge 2$, we have $$ \log n=\int_1^n\frac{dt}{t}\le\sum_{j=1}^{n-1}\frac{1}{j}\le 1+\int_1^{n-1}\frac{dt}{t}\le 1+\log n. $$ This in turn implies $$ ke^{-c}\frac{1}{n^c}\le a_n \le \frac{K}{n^C}. $$ Now, if $x=\frac{1}{e}$, then $\sum_n a_n =\infty$ follows from $a_n\ge ke^{-c}\frac{1}{n^c}$ for all but finitely many $n$. If $x=-\frac{1}{e}$, then $\frac{|a-\frac{n}{e}|^n}{n!}\to 0$ follows from $$\begin{eqnarray} \lim_{n\to\infty}\frac{|a-\frac{n}{e}|^n}{n!}&=& \lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{n!}\\ &=&e^{-ae}\lim_{n\to\infty}a_n\\ &\le&e^{-ae}\lim_{n\to\infty}\frac{K}{n^C}=0. \end{eqnarray}$$

  • $\begingroup$ Could you further elaborate on the part where you introduce c and C? $\endgroup$ – s0ulr3aper07 Jan 15 '19 at 22:06
  • $\begingroup$ @s0ulr3aper07 I hope this makes it clear ... $\endgroup$ – Song Jan 15 '19 at 22:37
  • $\begingroup$ Re: Stirling's Formula. For many problems like this, it suffices that $L=\lim_{n\to \infty} (n!)^{-1}(n/e)^n \sqrt n$ exists and is positive. That $L=1/\sqrt {2\pi}$ takes a lot more work to prove. $\endgroup$ – DanielWainfleet Jan 16 '19 at 21:48

When $x=1/e$ the numerator is equivalent (as $n\to \infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/\sqrt{2\pi n}$ and your series is therefore divergent.

  • 1
    $\begingroup$ And what happens if $x=-1/e$? ;-)) $\endgroup$ – Mark Viola Jan 15 '19 at 20:10

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