# Testing series for convergence/divergence

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?

• I think you will need Stirling formula. – 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 $$\begin{eqnarray} \lim_{n\to\infty}\frac{|a-\frac{n}{e}|^n}{n!}&=& \lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{n!}\\ &=&e^{-ae}\lim_{n\to\infty}a_n\\ &\le&e^{-ae}\lim_{n\to\infty}\frac{K}{n^C}=0. \end{eqnarray}$$

• Could you further elaborate on the part where you introduce c and C? – s0ulr3aper07 Jan 15 '19 at 22:06
• @s0ulr3aper07 I hope this makes it clear ... – Song Jan 15 '19 at 22:37
• 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. – 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.

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