Convergence of the series $\sqrt[n]n-1$ Let $a_n=\sqrt[n]n-1$. Does $\sum_{n=1}^\infty a_n$ converge?
Hint: $$\sqrt[n]n  = e^{\frac{\log n}{n}} &gt; 1+\frac{\log n}{n}$$
Note that for any $n\ge3$:
$$ n&gt;e$$
and so,
$$ n^{1/n} &gt; e^{1/n}$$

And since we know:
$$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ we can say:
$$ e^{1/n} = \sum_{k=0}^\infty \frac{1}{k!\;n^k} = 1 + \frac{1}{n} + \frac{1}{2n^2}+\cdots$$
Which, for $n\in\Bbb N$, is surely larger than the truncated series $1+1/n$, and so follows the strict inequality:
$$n^{1/n} &gt; e^{1/n} &gt; 1 + \frac{1}{n}$$

Subtract one from both sides, yielding:
$$\sqrt[n]{n}-1 = n^{1/n} - 1&gt;\frac1n$$

Which is true $\forall n\ge 3$. Evaluate the sum:
$$ \sum_{n\ge 3} (\sqrt[n]{n}-1) &gt; \sum_{n\ge 3} \frac{1}{n}$$ 

Thus, $\sum a_n$ is larger than the divergent harmonic sum, and is divergent too.