0
$\begingroup$

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 $a+b+c$.

$\endgroup$
3
  • 1
    $\begingroup$ 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
8
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.