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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!

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

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

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