TITLE: converge/divergence of a series with a n-th root
QUESTION [5 upvotes]: for the following series, prove for convergence/divergence 
  $$\sum\limits_{v=1}^\infty   \frac{\sqrt[n]{n}}{\sqrt[n]{n!}}$$

$$\sum\limits_{v=1}^\infty   \frac{\sqrt[n]{n}}{\sqrt[n]{n!}}=\sum\limits_{v=1}^\infty   \sqrt[n]\frac{n}{{n!}}=\sum\limits_{v=1}^\infty   \frac{1}{\sqrt[n]{(n-1)!}}$$
now 
$$\sum\limits_{v=1}^\infty   \frac{1}{{n}}=\sum\limits_{v=1}^\infty   \frac{1}{\sqrt[n]{n^n}}\leq\sum\limits_{v=1}^\infty   \frac{1}{\sqrt[n]{(n-1)!}}$$
$\sum\limits_{v=1}^\infty   \frac{1}{\sqrt[n]{n^n}}$ diverages so by the comparison test $\sum\limits_{v=1}^\infty   \frac{\sqrt[n]{n}}{\sqrt[n]{n!}}$ diverges too.
Is this proof valid?

REPLY [2 votes]: Sure.
Comparison test (statement):
Let $(a_n)_n$ and $(b_n)_n$ be two non-negative sequences of real numbers. Suppose that:
$1)$ $\sum_{n\ge1}a_n$ converges.
$2)$ $\exists n_0 \in \mathbb N$, such that $b_n \le a_n$, $\forall n \ge n_0$
Then, $\sum_{n\ge1} b_n$ converges.
Alternative form:
Let $(a_n)_n$ and $(b_n)_n$ be two non-negative sequences of real numbers. Suppose that:
$1)$ $\sum_{n\ge1} b_n$ diverges.
$2)$ $\exists n_0 \in \mathbb N$, such that $b_n \le a_n$, $\forall n \ge n_0$
Then, $\sum_{n\ge1}a_n$ diverges.