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?

  • $\begingroup$ 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 ? $\endgroup$ – Aryadeva 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
  • 2
    $\begingroup$ 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

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.)


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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.