Hint:
Let the general term be denoted by:
[tex]
a_k = \sqrt[k]{k} - 1[/tex]
Do you know how to prove that [itex]\lim_{k \rightarrow \infty}{a_k} = 0[/itex]?
Then, we have:
[tex]
k = (1 + a_k)^{k}[/tex]
Taking the same equality for [itex]k + 1[/itex], and subtracting this one, you ought to get:
[tex]
1 = (1 + a_{k + 1})^{k + 1} - (1 + a_k)^{k}[/tex]
Solve this equation for [itex]a_{k + 1}[/itex]. What do you get?
Because as [itex]k \rightarrow \infty[/itex], [itex]a_k[/itex] is an infinitesimal quantity, you may expand your expression for [itex]a_{k + 1}[/itex] in powers of [itex]a_{k}[/itex] 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 [itex]b_n[/itex], and the series [itex]\sum_{n = 1}^{\infty}{b_n}[/itex] 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.