How to show convergence of a particular series from first principles I am refreshing my techniques for series and am currently stuck with the following exercise:

Show that for every real number $x > 1$ the series 
  $$
\sum_k \frac{2^k}{1 + x^{2^k}}
$$
  converges.
  
  
*
  
*Hint: Add $\frac{1}{1-x}$. 
  

This exercise appears right at the start of the chapter so nothing other than the bare definition of convergence of a series (via convergence of the partical sums) can be used -- not even the convergence of the geometric series as this has not been established yet. Though basic sequence results are ok to be applied to the sequence of partial sums.
I can find a solution, but not via the hint and I was wondering how I could use the hint!
P.S. Here is the argument that I have:
Using the fact that $x > 1$ and $2^k \ge 2k$ for each $k \in \mathbb N$ We have
\begin{align}
s_n = \sum^n_{k=1} \frac{2^k}{1 + x^{2^k}} \le \sum^n_{k=1} \frac{2^k}{(1 + x)^{2^k}} \le \sum^n_{k=1} \frac{2^k}{(1 + x)^{2k}} \le \sum^n_{k=1} \frac{2^k}{2^{2k}} = \sum^n_{k=1} \left(\frac{1}{2}\right)^k
\end{align}
and we know that the right hand side converges as $n \to \infty$. This approach does not use the hint, and rests on the knowledge that $\sum_k (\frac{1}{2})^k < \infty$.
 A: Hint Let's add $\frac{1}{1-x}$ to the first term and the result to the next term:
$$\begin{align}\frac{1}{1-x}+\frac{1}{1+x}&&=&\frac{2}{1-x^2}\\
               \frac{2}{1-x^2}+\frac{2}{1+x^2} &&=&\frac{2^2}{1-x^{2^2}}\\
               \frac{2^2}{1-x^4}+\frac{2^2}{1+x^4} &&=&\frac{2^3}{1-x^{2^3}} \end{align}$$
Can you see the telescoping formula:
$$\frac{1}{1-x}+\left(\frac{2^0}{1+x^{2^0}}+\frac{2^1}{1+x^{2^1}}+\frac{2^2}{1+x^{2^2}}\right)=\frac{8}{1-x^8} $$
Telescoping formula is : ( pass the mouse over to see it) 

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

A: From 
$$
\left(1-x^{2^k} \right)\left(1+x^{2^k} \right)=1-x^{2^{k+1}} 
$$ you deduce, for $x^{2^k} \neq 1$,
$$
1+x^{2^k}=\frac{1-x^{2^{k+1}} }{1-x^{2^k}}
$$ then multiplying, factors telescope, giving

$$
\prod_{k=0}^n\left(1+x^{2^k} \right)=\frac{1-x^{2^{n+1}} }{1-x } \tag1
$$ 

Take the logarithm of the precedent finite product and differentiate, make $x \to x^{-1}$, you get

$$
\sum_{k=0}^n\frac{2^k}{1+x^{2^k}}=\frac{2^{n+1}}{x^{2^{n+1}}-1 }-\frac1{1-x }. \tag2
$$ 

