An infinite series Hopefully this question is not a duplication.
Consider the following infinite series:

$$\LARGE\sum_{k=0}^\infty\frac{2^k}{1+\frac{1}{x^{2^k}}}$$

We know the answer is $\frac{x}{1-x}$ if $|x|<1$. We also know the partial sum is actually given by
$\sum_{k=0}^n\frac{2^k}{1+\frac{1}{x^{2^k}}}=\frac{\sum_{k=1}^n kx^k}{\sum_{k=0}^n x^k}$. This formula can of course be proved by induction. What I want to know is a way to derive this formula. Since the bottom is the geometric series and the top is its derivative multiplied by $x$, there should be a nice way to derive it.
We know there is a way of finding the sum of the infinite series by appealing to double series, and this is the last thing I want to see.
Thanks a lot!
 A: To me, the sum can be derived informally as
$$\sum_{k=0}^\infty \frac{2^k x^{2^k}}{1+x^{2^k}}
= x \sum_{k=0}^\infty \frac{d}{dx}\left[ \log(1+x^{2^k}) \right]
= x \frac{d}{dx}\left[ \log\left(\prod_{k=0}^{\infty}1+x^{2^k}\right) \right]\\
= x \frac{d}{dx}\left[ \log\left(\prod_{k=0}^{\infty}\frac{1-x^{2^{k+1}}}{1-x^{2^k}}\right) \right]
= -x\frac{d}{dx}\left[\log(1-x)\right] = \frac{x}{1-x}
$$
The key is the summand in the original series is proportional to the derivative of logarithm of something simple. Once we turn the sum of log to a log of product and realize the product is a telescoping one, the rest is just following your nose.
A: Let $|x| < 1$ and $k \ge 0$. Using the factorization $1 - x^{2^{k+1}} = (1 - x^{2^k})(1 + x^{2^k})$, we find 
$$\frac{2^k x^{2^k}}{1 + x^{2^k}} = \frac{2^kx^{2^k}[(1 + x^{2^k}) - 2x^{2^k}]}{(1 + x^{2^k})(1 - x^{2^k})} = \frac{2^k x^{2^k}}{1 - x^{2^k}} - \frac{2^{k+1}x^{2^{k+1}}}{1 - x^{2^{k+1}}}.$$
So since $\lim\limits_{k\to \infty} \frac{2^k x^{2^k}}{1 - x^{2^k}} = 0$, the series telescopes to $\frac{x}{1 - x}$.
A: Let $f(x)$ be the series. Rewriting, we get \begin{eqnarray}f(x)=\sum_{k=0}^\infty\frac{2^kx^{2^k}}{1+x^{2^k}}\end{eqnarray}
Let $s_n$ be the $n$-th partial sum. We then have
\begin{align*}
\frac{x}{1-x}-s_n&=\frac{x}{1-x}-\frac{x}{1+x}-\cdots-\frac{2^nx^{2^n}}{1+x^{2^n}}\\
&=\frac{2^{n+1}x^{2^{n+1}}}{1+x^{2^{n+1}}}
\end{align*}
One can show that the above tends to 0 as $n\to \infty$, for $|x|<1$. 
