<s>[INST] $\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows      $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms.   The sum is to be expressed in terms of $x$ and $n$. I tried setting a formula for the $n$-th term and setting it into a difference but ran into a dead end.
 [/INST] A supplement to the nice answer from @labbhattacharjee. We see an example of telescoping based upon
\begin{align*}
\color{blue}{\frac{2^k}{1-x^{2^k}}+\frac{2^k}{1+x^{2^k}}}
=\frac{2^k\left(1+x^{2^k}\right)+2^k\left(1-x^{2^k}\right)}{\left(1-x^{2^k}\right)\left(1+x^{2^k}\right)}
&\color{blue}{=\frac{2^{k+1}}{1-x^{2^{k+1}}}}\tag{1}
\end{align*}

We obtain according to (1)
  \begin{align*}
\color{blue}{\sum_{k=0}^n\frac{2^k}{1+x^{2^k}}}&=\sum_{k=0}^n\left(\frac{2^{k+1}}{1-x^{2^{k+1}}}-\frac{2^k}{1-x^{2^k}}\right)\\
&=\sum_{k=0}^n\frac{2^{k+1}}{1-x^{2^{k+1}}}-\sum_{k=0}^n\frac{2^k}{1-x^{2^k}}\\
&=\sum_{k=1}^{n+1}\frac{2^{k}}{1-x^{2^{k}}}-\sum_{k=0}^n\frac{2^k}{1-x^{2^k}}\tag{2}\\
&\color{blue}{=\frac{2^{n+1}}{1-x^{2^{n+1}}}-\frac{1}{1-x}}\tag{3}
\end{align*}

Comment:


*

*In (2) we shift the index of the left-hand sum to start with $k=1$.

*In (3) we do the telescoping.
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