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 
 A: 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.
A: Hint: Try some small cases ($n=1,2,3$). The pattern should become clear, and you can most likely prove it by induction on $n$.
A: Hint: For $x\ne 1$, add $\frac{-1}{x-1}$ in front, and observe the telescoping.  
A: note: having just seen Andre's lovely solution this is very much a hammer-and-chisel job! however there was some devil in the detail, and i did learn something from the exercise!
define:
$$
D_m=\prod_{k=1}^m (1+x^{2^{m-1}}) = \sum_{k=0}^{2^m-1}x^k
$$
and
$$
N_m = \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} \tag{1}
$$
and now define the sum 
$$
S_m=\frac{N_m}{D_m}
$$
then
$$
S_m+\frac{2^m}{x^{2^m}+1} = \frac{(x^{2^m}+1)N_m + 2^mD_m}{D_{m+1}}
$$
from (1) we have
$$
N_{m+1} = \sum_{k=1}^{2^{m+1}-1} (2^{m+1}-k)x^{k-1} \\
=  \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} +\sum_{k=1}^{2^m-1} 2^m x^{k-1} +  \sum_{k=2^m}^{2^{m+1}-1} (2^{m+1}-k)x^{k-1} \\
= N_m + 2^m(D_m -x^{2^m-1}) + \sum_{k=0}^{2^m-1} (2^m-k)x^{2^m+k-1} \\
= (1+x^{2^m}) N_m + 2^m D_m
$$
hence, with the obvious definition of $S_m$ we have the inductive step:
$$
S_m = \frac{N_m}{D_m} \Rightarrow S_{m+1} = \frac{N_{m+1}}{D_{m+1}}
$$
