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Show that for all non-negatives integers $n$, it is true that $$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$$ For $x \ne 1$, $x \ne -1$ I tried to solved it by appliying geometric series but I got the following after doing the $2f\left(x\right)-f\left(x\right)$ step, where $f\left(x\right)=\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}$

$$f\left(x\right)=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n}}}$$ Not sure if Fermat's theorem is needed in order to solve it

Diego Gallo
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  • The case $n=0$ is $\tfrac{1}{1+x}=\tfrac{1}{x-1}+\tfrac{2}{1+x^2}$, so you must have a typo somewhere. Can you spot it? – J.G. Nov 06 '20 at 23:51
  • I think the RHS should be $\dfrac 1{x-1}+\dfrac{2^{n+1}}{1\color{red}{-}x^{2^{n+1}}}$ – Prasun Biswas Nov 06 '20 at 23:54
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    Do not write $x!=-1$ That means the factorial of $x$ is $-1$. Instead write $x \ne -1$ In TeX, write x \ne 1 – GEdgar Nov 07 '20 at 00:45

2 Answers2

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Subtract $1/(x-1)$ from LHS and note that,

$$\frac 1{x+1}-\frac 1{x-1}=-\frac 2{x^2-1} \\ \frac 2{x^2+1}-\frac 2{x^2-1}=-\frac 4{x^4-1}\\ \frac 4{x^4+1}-\frac 4{x^4-1}=-\frac 8{x^8-1}$$

and so on. Do you see the telescoping pattern?

To finish, note that,

$$\frac {2^n}{x^{2^n}+1}-\frac {2^n}{x^{2^n}-1}=-\frac{2^{n+1}}{x^{2^{n+1}}-1}=\frac{2^{n+1}}{1-x^{2^{n+1}}}$$

Prasun Biswas
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A proof by induction is elementary and not difficult. Let $$S_n(x) = \sum_{k=0}^n \frac{2^k}{1 + x^{2^k}}, \quad P_n(x) = \frac{1}{x-1} + \frac{2^{n+1}}{\color{red}{1 - x^{2^{n+1}}}}.$$ Note that you have a typographical error for $P_n(x)$; the correct expression is shown in red text. The claim is that for all nonnegative integers $n$, we have $S_n = P_n$. The proof of the inductive step is simply

$$\begin{align} S_{n+1}(x) &= S_n(x) + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\ &= P_n(x) + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\ &= \frac{1}{x-1} + \frac{2^{n+1}}{1 - x^{2^{n+1}}} + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\ &= \frac{1}{x-1} + 2^{n+1} \left( \frac{(1 + x^{2^{n+1}}) + (1 - x^{2^{n+1}})}{(1 - x^{2^{n+1}})(1 + x^{2^{n+1}})} \right) \\ &= \frac{1}{x-1} + 2^{n+1} \left( \frac{2}{1 - x^{2^{n+2}}} \right) \\ &= P_{n+1}(x). \end{align}$$

The way that this simplifies also suggests that a direct proof is possible via algebraic simplification of a telescoping sum, e.g., $$\frac{2^{k+1}}{1 - x^{2^{k+1}}} = \frac{2^k}{1 - x^{2^k}} + \frac{2^k}{1 + x^{2^k}}$$ suggests defining $$a_k (x) = \frac{2^k}{1 + x^{2^k}}, \quad b_k(x) = \frac{2^k}{1 - x^{2^k}} $$ hence $$a_k(x) = b_{k+1}(x) - b_k(x)$$ and $$S_n(x) = \sum_{k=0}^n a_k(x) = \sum_{k=0}^n b_{k+1}(x) - b_k(x) = b_{n+1}(x) - b_0(x) = P_n(x).$$

heropup
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