$\sum_{k=1}^{\infty} \frac 1 {k^{1+\frac 1 k}} $
What I've tried:
$a = $ $\sum_{k=1}^{\infty} \frac 1 {k^{1+\frac 1 k}} $
$b = $ $\frac 1 k$
Limit comparison test:
$\lim_{n\to {\infty}} \frac a b = 1$
Therefore, by Limit comparison test, a and b diverge or converge together.
Because b diverges (p-series), a must also diverge.
Have I made any mistakes?
Yes you are absolutely right indeed
$$\frac{\frac 1 {k^{1+\frac 1 k}}}{\frac1k}=\frac k {k^{1+\frac 1 k}}=\frac 1 {k^{\frac 1 k}}\to 1$$
and by limit comparison test we can conclude.