For which $a>0$ does $\sum_{k=1}^{\infty} \frac{1}{k^a+a^{-k}}$ converge? 
For which $a$ does $$\sum_{k=1}^{\infty} \frac{1}{k^a+a^{-k}}$$ converge?

So far, I have figured out (and I hope I'm not wrong about this) that the series converges for $a > 1$ since $\frac{1}{k^a}$ converges for $a > 1$ and $$\frac{1}{k^a+a^{-k}} < \frac{1}{k^a}$$ for $a > 1$.
The book says it also converges for $a < 1$ and specifies that it diverges for $a=1$ but I don't know how to reach that conclusion, and could use some help.
Update: Forgot to mention, $a > 0$, sorry!
 A: I am assuming $a > 0$, which is implicit in your question.


*

*For $a=1$, you have $$
\frac{1}{k^a+a^{-k}} = \frac{1}{k+1}$$
so by comparison with the Harmonic series the series
$$
\sum_{k=1}^n \frac{1}{k^a+a^{-k}} 
$$
diverges.

*For $0 < a < 1$, as you noted
$$
\frac{1}{k^a+a^{-k}} = \frac{a^k}{a^kk^a + 1} = a^k b_k$$
where $b_k = \frac{a^k}{a^kk^a + 1} \xrightarrow[k\to \infty]{}1$
so by comparison (all is non-negative) with the series of general term $a^k$ the series converges.

*For $a > 1$, as you noted
$$
\frac{1}{k^a+a^{-k}} = \frac{1}{k^a}\frac{1}{1+ \frac{1}{k^a a^k}}$$
so since $\frac{1}{1+ \frac{1}{k^a a^k}}\xrightarrow[k\to\infty]{} 1$ by comparison with the series of general term $\frac{1}{k^a}$ (which is a $p$-series, and $a > 1$) the series converges.
A: Hint:
As this is a series with positive terms, the simplest is to use equivalents:


*

*if $\,a>1$, $\,k^a+a^{-k}\sim_\infty k^a$, hence $\;\dfrac1{k^a+a^{-k}}\sim_\infty \dfrac1{k^a}$, which converges.

*if $\,0<a<1$, $\,k^a+a^{-k}\sim_\infty a^{-k}$, hence $\;\dfrac1{k^a+a^{-k}}\sim_\infty  a^k$, which converges.

*if $a=1$, we have the (shifted) harmonic series, which diverges.

A: $$\frac{1}{k^a+a^{-k}} < \frac{1}{a^{-k}}=a^k.$$
If $0<a<1$, $\sum a^k$ is a convergent geometric series.If $a=1$ the series is
$$\sum_{k=1}^\infty\frac{1}{k+1},$$
which is divergent.
