Hint:
Let the general term be denoted by:
<br />
a_k = \sqrt[k]{k} - 1<br />
Do you know how to prove that \lim_{k \rightarrow \infty}{a_k} = 0?
Then, we have:
<br />
k = (1 + a_k)^{k}<br />
Taking the same equality for k + 1, and subtracting this one, you ought to get:
<br />
1 = (1 + a_{k + 1})^{k + 1} - (1 + a_k)^{k}<br />
Solve this equation for a_{k + 1}. What do you get?
Because as k \rightarrow \infty, a_k is an infinitesimal quantity, you may expand your expression for a_{k + 1} in powers of a_{k} up to the first non-vanishing order. What do you get?
The solution for this asymptotic recursion relation would give you a comparison general term b_n, and the series \sum_{n = 1}^{\infty}{b_n} is pretty easy to test for convergence. Then, you may use the
[STRIKE]Ratio comparison test[/STRIKE].
EDIT:
Use the Asymptotic comparison test mentioned
here instead.