# 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!

• It does diverge for $a=1$. – Kushal Bhuyan Jun 3 '16 at 12:59

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.

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.

$$\frac{1}{k^a+a^{-k}} < \frac{1}{a^{-k}}=a^k.$$ 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.