A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}$, where $a$, $b$, and $c$ are integers. Find $a+b+c$.

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    $\begingroup$ Post your thoughts and stuff you have tried. Trying with small values of $n$ will probably help you. $\endgroup$ Jan 25 '15 at 18:49
  • $\begingroup$ But I just have no clue where to start. Probably a sigma? But induction would be really helpful. But its just that I haven't mastered induction and I need a detailed explanation. $\endgroup$
    – Roderick
    Jan 25 '15 at 18:51
  • $\begingroup$ math.stackexchange.com/questions/1038955/… is kinda similar. $\endgroup$
    – r9m
    Jan 26 '15 at 3:58

Note that

$$\frac{2^{k+1}}{3^{2^{k+1}}-1} - \frac{2^{k}}{3^{2^{k}}-1} = \frac{2^{k}}{3^{2^{k}}-1} \left ( \frac{2}{3^{2^{k}}+1} - 1\right ) = -\frac{2^k}{3^{2^{k}}+1}$$

Thus the finite sum is a telescoping sum and depends only on the end terms of the series.


Manuel Lafond’s suggestion that you try some small values of $n$ is always worth pursuing if you don’t see a cleverer approach. By actual calculation you can fairly readily discover that $$\begin{align*} S_0&=\frac12=1-\frac12\;,\\ S_1&=\frac9{10}=1-\frac1{10}\;,\\ S_2&=\frac{409}{410}=1-\frac1{410}\;,\text{ and}\\ S_3&=\frac{1345209}{1345210}=1-\frac1{1345210}\;. \end{align*}$$ We’d like to find $a,b$, and $c$ such that $$\begin{align*} \frac12&=\frac{a^b}{3^{2^c}-1}\;,\\ \frac1{10}&=\frac{a^{1+b}}{3^{2^{1+c}}-1}\;,\\ \frac1{410}&=\frac{a^{2+b}}{3^{2^{2+c}}-1}\;,\text{ and}\\ \frac1{1345210}&=\frac{a^{3+b}}{3^{2^{3+c}}-1}\;. \end{align*}$$

The third one looks like a good place to start: the fourth involves uncomfortably large numbers, and the first and second look as if they might allow too many reasonable possibilities. For the third we need $2^{2+c}$ to be a number $m$ such that $3^m-1$ is a multiple of $410$. The first possibility is $m=8$, and indeed $3^8-1=6560=16\cdot410$. That suggests that we try the hypothesis that $$S_n=1-\frac{2^{n+2}}{3^{2^{n+1}}-1}\;.$$

This yields the correct values for $n=0,1$, and $3$, so it’s almost certainly correct, and all that remains is to prove it by induction, as you suggested. The induction step requires showing that $$1-\frac{2^{n+2}}{3^{2^{n+1}}-1}+\frac{2^{n+2}}{3^{2^{n+1}}+1}=1-\frac{2^{n+3}}{3^{2^{n+2}}-1}\;,$$ which is a straightforward bit of algebra.

  • $\begingroup$ I'm sorry to say that your answer's incorrect, do you know where you've miscalculated? $\endgroup$
    – user359548
    Aug 12 '16 at 23:10
  • $\begingroup$ @MathMuse: It is correct for $n=0$: $$1-\frac{2^{0+2}}{3^{2^{0+1}}-1}=1-\frac{4}{9-1}=\frac{1}2=S_0\;.$$ $\endgroup$ Aug 12 '16 at 23:13
  • $\begingroup$ Oh, I see. Sorry, I was looking at the last part of the equation. +1 $\endgroup$
    – user359548
    Aug 12 '16 at 23:16
  • $\begingroup$ @MathMuse: No, I wrote $$S_n=\frac{2^{n+2}}{3^{2^{n+1}}-1}\;.$$ $\endgroup$ Aug 12 '16 at 23:18
  • $\begingroup$ I edited my comment. $\endgroup$
    – user359548
    Aug 12 '16 at 23:20

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