Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Concerning this question of mine which involves telescoping in the solution, I was wondering if it is possible to express




as infinite telescoping products.

Is it possible?



The infinite product $$\ln(x)=(x-1)\prod_{k=1}^{\infty}\frac{2}{x^{2^{-k}}+1}$$

given bellow by @Zarrax, was first published by Ludwig von Seidel. See Theorem 4 in "The Logarithmic Constant: log(2)" by Xavier Gourdon and Pascal Sebah.

share|cite|improve this question
I just found an infinite product for $e^x$ due to Jonathan Sondow and Jesús Guillera $$\displaystyle e^x=\prod_{n=1}^{\infty}\left( \prod_{k=0}^{n}(kx+1)^{(-1)^{k+1}\binom{n}{k}}\right)$$ – Neves Dec 31 '10 at 12:17
up vote 6 down vote accepted

Let $A_n(x) = n(x^{1 \over n} - 1)$. Then if $B_n$ denotes $A_{2^{n}}$, then $B_n(x) = 2^n(x^{1 \over 2^n} - 1)$, and $\lim_{n \rightarrow \infty} B_n(x)$ equals $\ln(x)$ because $B_n(x)$ is a subsequence of $A_n(x)$. Note that $${B_n(x) \over B_{n-1}(x)} = {2^n(x^{1 \over 2^n} - 1) \over 2^{n-1}(x^{1 \over 2^{n-1}} - 1)}$$ $$= {2 (x^{1 \over 2^n} - 1) \over (x^{1 \over 2^n} - 1)(x^{1 \over 2^n} + 1)}$$ $$= {2 \over x^{1 \over 2^n} + 1}$$ So $B_n$ can be written as a finite telescoping product $$B_n = (x- 1)\prod_{i = 1}^{n} {2 \over x^{1 \over 2^i} + 1}$$ So taking the limits as $n$ goes to $\infty$ one has $$\ln(x) = (x-1)\prod_{i = 1}^{\infty} {2 \over x^{1 \over 2^i} + 1}$$ So at least a subsequence of the partial products gives a telescoping product.

share|cite|improve this answer
This is great! Any $a^i$ will work as long $a>1$. – Pedro Tamaroff Apr 5 '12 at 18:52

This is a comment on generalities(therefore just community wiki).

For infinite products of expressions of the form $(1 + \mathrm{something})$, it is often more productive to take the logarithm and convert it into an infinite sum.

There is a little bit of an issue of convergence here. If one of the expressions is $0$, then there would be trouble, etc.. A precise handling of such stuff can be found in the book of Ahlfors on Complex Analysis.

share|cite|improve this answer

Well, we could produce a very contrived example for $e^x$ by defining $S_m(x) = \sum_{k=0}^m x^k/k! $ and let $$ P_N(x) = \prod_{k=0}^N \frac{S_{k+1}(x)}{S_k(x)} = S_{N+1}(x).$$

And so $\lim_{N \rightarrow \infty} P_N(x) = e^x .$

More concretely, $e^x$ cannot be represented as an infinite product, such as we can do for $\sin x$ for example, because any zeros of the factors would also be zeros of $e^x$ but $e^x$ is positive for all $x \in \mathbb{R}.$

EDIT: Here's a slightly less contrived product formula for $e^x .$


$$ P_n(x) = \prod_{k=1}^n \left( 1+ \frac{x}{n} \right)^{n/k(k+1)} $$

then $e^x = \lim_{n \rightarrow \infty} P_n(x),$ although there's no telescoping going on here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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