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The series is given by :

$$ \sum_{n \geq 1} \frac 1 {n!}\left(\frac n e\right)^n$$

I tried to test its convergence by using a variant of Raabe's test as shown in Bartle's "Introduction to Real Analysis" book.

However, the limit that I got is $-\infty$, which I think fails Raabe's test (?)

I also tried using comparison test, limit comparison test, ratio test, which all failed.

Could anyone help me determine whether this series converges or not?

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  • $\begingroup$ So essentially there is a $n^n$ in the numerator? $\endgroup$ – imranfat Jan 31 '18 at 2:21
  • $\begingroup$ Is it an infinite sum ? Jonelle ? $\endgroup$ – Isham Jan 31 '18 at 2:24
  • $\begingroup$ yes, this is an inifinite sum, and yes there is an n^n in the numerator $\endgroup$ – Jonelle Yu Jan 31 '18 at 2:47
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    $\begingroup$ If it's a series, then write it as a series, not a sequence. $\endgroup$ – zhw. Jan 31 '18 at 4:39
  • $\begingroup$ This is easy with Stirling. $\endgroup$ – zhw. Jan 31 '18 at 19:09
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The term test is also inconclusive since $(n/e)^n/n! \to 0$ as $n \to \infty$. However, you can quickly establish divergence of the series since Stirling's approximation gives $(n/e)^n/n! \sim C/\sqrt{n}$.

If you are interested, Raabe's test states that given a series $\sum a_n$ with positive terms, if we obtain

$$\tag{*}\lim_{n\to \infty}\left(n \frac{a_n}{a_{n+1}} - n - 1 \right) = r,$$

then the series converges if $r > 0$ and diverges if $r < 0$. The case where $r = 0$ is inconclusive.

In this case, $a_n = (n/e)^n/n!$, so

$$\frac{a_n}{a_{n+1}} = \frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}\frac{n^n}{n!e^n} = \frac{e}{(1 + 1/n)^n},$$ and

$$n \frac{a_n}{a_{n+1}} - n - 1 = n\left(\frac{e}{(1+1/n)^n}- 1 \right) - 1.$$

Since $(1 + 1/n)^n \to e$, the term in parentheses converges to $0$. Also, this shows why the ratio and root test are inconclusive -- since $a_{n+1}/a_n \to 1$.

It can be shown that

$$\frac{e}{2n +2} < e - (1 + 1/n)^n < \frac{e}{2n+1}.$$

(A proof of this inequality is given in Problems and Theorems in Analysis I by Polya and Szego.)

Hence,

$$\frac{e}{(1+1/n)^n} \frac{n}{2n+2} - 1< n\left(\frac{e}{(1+1/n)^n}- 1 \right) - 1 < \frac{e}{(1+1/n)^n} \frac{n}{2n+1} - 1.$$

Since the limits of the left-hand and right-hand sides are both $-1/2$, it follows by the squeeze theorem that the limit in (*) is $r = -1/2$ and we can conclude that the series diverges.

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  • $\begingroup$ wow, this is smart, especially the inequality you introduced to come up with the limit of the expression $\endgroup$ – Jonelle Yu Jan 31 '18 at 7:57
  • $\begingroup$ @JonelleYu: You're welcome. I'll edit your post to add $\sum$ to show this is a series question, This seems to have caused confusion although you state explicitly series. $\endgroup$ – RRL Jan 31 '18 at 18:00

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