Find the Sum of the Series: $1/(x+1) + 2/(x^2 + 1) + 4/(x^4 +1) +\cdots$ $n$ terms Find the sum of $n$ terms the following series:
$$\frac1{x+1} + \frac2{x^2 + 1} + \frac4{x^4 +1} +\cdots\qquad n\text{ terms}$$
$t_n$ seems to be $\dfrac{2^{n-1}}{x^{2^{n-1}} + 1}$
But after that I'm not sure as to how to proceed 

Just another approach.
Let $x$ be a real number such that $|x|>1$. Observe that
$$
1+x^{2^{n}}=\frac{x^{2^{n+1}}-1}{x^{2^{n}}-1} \quad n=0,1,2,\ldots,
$$ giving
$$
\prod_{n=0}^{N} \left(1+x^{2^{n}}\right)=\frac{x^{2^{N+1}}-1}{x-1}
$$
Applying the logarithmic function gives 
$$
\sum_{n=0}^{N}\log \left(1+x^{2^{n}}\right)=\log\left(x^{2^{N+1}}-1\right)-\log (x-1)
$$
differentiating with respect to $x$
$$
\sum_{n=0}^{N}\frac{2^{n}x^{2^{n}-1}}{x^{2^{n}}+1}=\frac{2^{N+1}x^{2^{N+1}-1}}{x^{2^{N+1}}-1}-\frac{1}{x-1}
$$ equivalently 
$$
\sum_{n=0}^{N}\frac{2^{n}}{x^{2^{n}}+1}=\frac{1}{x-1}-\frac{2^{N+1}}{x^{2^{N+1}}-1}
$$
and then making $N$ tend to $+\infty$ leads to a closed form for your series.
