Hopefully this question is not a duplication.

Consider the following infinite series:


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!

  • 1
    $\begingroup$ There is a closed form for $\frac{\sum_{k=1}^nkx^k}{\sum_{k=1}^nx^k}$, I believe. $\endgroup$ – Akiva Weinberger May 15 '15 at 9:31

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.

  • $\begingroup$ Thanks. I guess I never thought of "log of product" and telescoping thing. I will do the finite sum (and finite product) and then the differentiation. I know if certain thing is uniformly convergent, then infinite summation and differentiation can be interchanged (I am recalling from my head, I don't know the exact formulation). Anyway this condition should be verified in this case, but I don't bother to check. $\endgroup$ – GRR May 15 '15 at 8:32
  • $\begingroup$ By telescoping I meant the third equality is great so that we can use telescoping. $\endgroup$ – GRR May 15 '15 at 8:40
  • $\begingroup$ This works for the finite sum, too, right? $\endgroup$ – Akiva Weinberger May 15 '15 at 9:30
  • $\begingroup$ @columbus8myhw Yes, this works for finite sum. In fact, if one want to prove the above expression rigorously, One use this trick to derive the correct expression for finite sum and then take the limit. $\endgroup$ – achille hui May 15 '15 at 10:03
  • $\begingroup$ @achillehui Is that really necessary? Are there examples where the use of the "informal" logic of your answer give the wrong result? $\endgroup$ – Akiva Weinberger May 15 '15 at 10:05

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 $\frac{x}{1 - x}$.


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 $|x|<1$.

  • $\begingroup$ Um....$\frac{2x^2}{1-x^2}+\frac{2x^2}{1+x^2}$ seems to be $\frac{4x^2}{1-x^4}$....... $\endgroup$ – GRR May 15 '15 at 8:21
  • $\begingroup$ Thanks for pointing out the mistake in my first attempt. I've corrected it. $\endgroup$ – Alex Fok May 15 '15 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.