Convergence of sequence and sum.

Question

Do the sequence $$\frac{n}{{n!}^{\frac{1}{n}}}$$ and the sum

$$\sum_{n=1}^\infty \frac{n}{{n!}^{\frac{1}{n}}}$$ converge? And if they do converge then to what real number. It seems to me that the sequence converges but I have no clue about the sum.

Answer 1

We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see this question).

We have that $$ \lim_{n\to\infty}\biggl(\frac{n^n}{n!}\biggr)^{1/n}=\lim_{n\to\infty}\frac{(n+1)^{n+1}n!}{(n+1)!n^n}=\lim_{n\to\infty}\biggl(1+\frac1{n}\biggr)^n=e. $$ Hence, the sequence converges but the series diverges.

Answer 2

Regarding just the sum, since you say you have a handle on the term by itself.

For $n>1$, we have $1 < n, 2 < n, \dots, n-1 < n$ and $n = n$, so $n! < n^n$. Then $n!^{1/n} < (n^n)^{1/n} = n$. So, for $n>1$,

$$ \frac{n}{n!^{1/n}} > \frac{n}{n} = 1 $$

and the sum cannot converge because the terms do not go to zero.

Did you think the limiting value of the term was zero?

Answer 3

Taking the logarithm,

$$\log s_n=-\frac1n\sum_{k=1}^n\log\frac kn.$$

In the limit,

$$\log s_\infty=-\int_0^1\log x\,dx=1.$$