How to prove divergence of this series Evaluate $\displaystyle \lim_{n\rightarrow \infty }(\sqrt[n]{n}-1)^n$  prove that  $\displaystyle  \sum_{n=1}^{\infty }(\sqrt[n]{n}-1)$ diverges.

(Hint : use comparison test and $\displaystyle \sqrt[n]{n}=e^{\ln n/n}=\sum_{k=0}^{\infty }\frac{1}{k!}(\frac{\ln n}{n})^{k}$  without using l'Hôpital's rule.
I tried this
let y = $\displaystyle \lim_{n\rightarrow \infty }(\sqrt[n]{n}-1)^n$ then 
$lny = ln \displaystyle \lim_{n\rightarrow \infty }(\sqrt[n]{n}-1)^n$
$lny = \displaystyle \lim_{n\rightarrow \infty }nln(\sqrt[n]{n}-1)$ 
$lny = \displaystyle \lim_{n\rightarrow \infty }n \times \lim_{n\rightarrow \infty } ln(\sqrt[n]{n}-1)$
$lny = \displaystyle \lim_{n\rightarrow \infty }n \times (ln \lim_{n\rightarrow \infty } (\sqrt[n]{n}-1))$
I know that $\displaystyle \lim_{n\rightarrow \infty } (\sqrt[n]{n}) = 1$ so,$lny = \displaystyle \lim_{n\rightarrow \infty }n \times 0 = 0 $ Therefore $lny = 0 -> y=1$ 
$\displaystyle \lim_{n\rightarrow \infty }(\sqrt[n]{n}-1)^n = 1$
is this right?
 A: Hint for showing series divergence:
Define $a_n = n^{1/n}-1$.  Then $\ln n = n \ln(1+a_n) \leqslant na_n$, and
$$\frac{\ln n}{n} \leqslant a_n= n^{1/n} -1.$$
Apply the binomial theorem to show that sequence converges to $0$ (without using L'Hospital's rule).
Note that for $n \geqslant 2$,
$$n^{1/n} = 1 + a_n \geqslant 1 \implies n = (1 + a_n)^n \geqslant \frac{n(n-1)}{2}a_n^2 \implies a_n^2 \leqslant \frac{2}{n-1}.$$
Hence,
$$0 \leqslant a_n \leqslant \frac{2^{1/2}}{(n-1)^{1/2}},$$
and for $n \geqslant 4$
$$0 \leqslant (n^{1/n}-1)^n \leqslant \frac{2^{n/2}}{(n-1)^{n/2}} < \left(\frac{2}{3}\right)^{n/2}.$$
Taking the limit and applying the squeeze principle we find $(n^{1/n}-1)^n \to 0$.
A: By the binomial theorem,
for $n \ge 2$,
$(1+1/n)^n
=\sum_{i=0}^n \binom{n}{i}\frac1{n^i}
=\sum_{i=0}^n \frac{\prod_{j=0}^{i-1} (n-j)}{i!n^i}
\le \sum_{i=0}^n \frac{1}{i!}
< e
$,
so
$1+1/n < e^{1/n}
< n^{1/n}$
for $n \ge 3$.
Therefore,
$n^{1/n}-1
> 1/n$
for $n \ge 3$,
so
$\sum_{n \ge 1} (n^{1/n}-1)
$ grows 
like $\sum 1/n$
at least,
and so diverges.
Note that
the key inequality is
$e^{1/n}
< n^{1/n}
$ for $n \ge 3$.
Ridiculous, isn't it?
