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Let $a \in (1, \infty)$. Calculate $$ \lim\limits_{n \to \infty} \sum_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $$

Here's what I tried:

Let $ x_n = \sum_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $. Let $a_n, b_n$ so that $a_n < x_n < b_n$. If $a_n \to l, b_n \to l$ with $ l \in \mathbb{R}$ then $x_n \to l$. I tried saying that $$ {{1+2+2^2+\dots+2^{n-1}}\over{a^{2n-1}+1}} < x_n < {{1+2+2^2+\dots+2^{n-1}}\over{a+1}} $$ but the limit of the lefthand-side fraction is $0$ and the one for the righthand-side is $\infty$. How can I solve this?

Jimmy R.
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Victor
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  • Is it $2^{n-1}$ or $2^{k-1}$ in the numerator? – Did Nov 05 '16 at 08:49
  • @Did, it's $2^{k-1}$, sorry! – Victor Nov 05 '16 at 08:51
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    Then the limit is obviously finite for every $a>\sqrt2$ (but the exact value of the sum probably has no closed form) and infinite for every $a\leqslant\sqrt2$. For the former statement, use $$\frac{2^{k-1}}{a^{2k-1}+1}\leqslant\frac{2^{k-1}}{a^{2k-1}}= \frac1a\left(\frac{2}{a^2}\right)^{k-1}$$ For the latter statement, note that, when $a\leqslant\sqrt2$, the general term does not go to $0$. – Did Nov 05 '16 at 08:53
  • @Did, can you elaborate an answer, please? I would really like to understand the exercise – Victor Nov 05 '16 at 08:56
  • "Elaborated" in my comment. – Did Nov 05 '16 at 08:57
  • What was more "elaborated" in the (readily accepted) answer below than in my comment? Please be specific. – Did Nov 05 '16 at 11:41

2 Answers2

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With equivalents: $$\frac{2^{k-1}}{a^{2k-1}+1}\sim_\infty\frac{2^{k-1}}{a^{2k-1}}=\frac a2\biggl(\frac2{a^2}\biggr)^k,$$ and the latter converges if and only if $\dfrac2{a^2}<1\iff (1<)\, a<\sqrt2$.

Bernard
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Let $a>\sqrt2$, we have that \begin{align} \sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}&=\frac a{2a+2}+\sum_{k=0}^{\infty} \frac{2^k}{a^{2k+1}}(1+a^{-1-2k})^{-1}\\ &=\frac a{2a+2}+\sum_{k=0}^{\infty} \frac{2^k}{a^{2k+1}}\sum_{n=0}^{\infty}(-a^{-1-2k})^n \end{align} Applying Fubini's theorem, \begin{align} \sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1} &=\frac a{2a+2}+\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}\sum_{k=0}^{\infty}(\frac2{a^{2n+2}})^k\\ &=\frac a{2a+2}+\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}(1-\frac2{a^{2n+2}})^{-1}\\ &=\frac a{2a+2}+\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}-\frac2{a^{n+1}}}\\ &=\frac a{2a+2}-\sum_{n=1}^{\infty}(-1)^n\frac{a^n}{a^{2n}-2}\tag{*}\label{*}\\ &=\frac a{2a+2}-\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}(1+\frac2{a^{2n}-2})\\ &=\frac a{2a+2}+\frac1{a+1}-\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}\frac2{a^{2n}-2}\\ &=\frac 1 2+\frac1 {2a+2}-\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}\frac2{a^{2n}-2} \end{align} This formula can give an approximation while $a\to\infty$.
We can see that the sum $\to \infty$ when $a\to \sqrt2_+$ : $$\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}=\frac 1 2+\frac1 {2a+2}+\bbox[5px,border:2px solid red]{\frac2{a(a^2-2)}}-\sum_{n=2}^{\infty}\frac{(-1)^n}{a^n}\frac2{a^{2n}-2}$$ which gives : $$\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}\operatorname*{\sim}_{a\to\sqrt2_+}\frac1{2(a-\sqrt2)}$$

More exactly, from $\eqref{*}$, \begin{align} \sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1} &=\frac a{2a+2}-\frac12\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n+\sqrt2}-\frac12\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n-\sqrt2}\\ &=\frac a{2a+2}+\frac12\sum_{n=1}^{\infty}\frac{1}{a^n+\sqrt2}+\frac12\sum_{n=1}^{\infty}\frac{1}{a^n-\sqrt2}-\sum_{n=1}^{\infty}\frac{1}{a^{2n}+\sqrt2}-\sum_{n=1}^{\infty}\frac{1}{a^{2n}-\sqrt2} \end{align} One can then express this ugly sum using some uglier q-psi functions.

Aforest
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