$\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}$ Divergent? Is this thorough enough?  

$$\frac{1}{n^{1+\frac{1}{n}}} = \frac{1}{n \cdot n^{\frac{1}{n}}} \ge \frac{1}{n} \quad as\ n\to \infty$$

Therefore by comparison w/ $\frac{1}{n}$, $\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}$ is divergent. 

Thanks!
What you have written is incorrect, since it is not true that $\frac{1}{n \cdot n^{\frac{1}{n}}} \ge \frac{1}{n}$.  Indeed, $\frac{1}{n \cdot n^{\frac{1}{n}}} &lt; \frac{1}{n}$ for all $n&gt;1$.  You seem to be trying to claim that your inequality holds "in the limit", but it is unclear what you mean by this precisely or how you are claiming this would imply your sum diverges.

To solve the problem correctly, I would suggest trying to compare $\frac{1}{n \cdot n^{\frac{1}{n}}}$ to $\frac{1}{2n}$ instead.
$\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges if and only if $s &lt; 1$. Clearly you have that $1+ \frac{1}{n} \geq 1$ and for $n\to\infty$ you have that $1+ \frac{1}{n} \to 1$