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

$$\ln(x)=\lim_{n\to\infty}n\left(x^{\frac1{n}}-1\right)$$

and

$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$

as infinite telescoping products.

Is it possible?

Thanks.


EDIT:

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.

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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
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3 Answers 3

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.

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

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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 $e^x .$

Let

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

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

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