Joan Thompson

2022-01-04

Consider the following infinite series:

$\sum _{k=0}^{\mathrm{\infty}}\frac{{2}^{k}}{1+\frac{1}{{x}^{2k}}}$

Hattie Schaeffer

Beginner2022-01-05Added 37 answers

To me, the sum can be derived informally as

$\sum _{k=0}^{\mathrm{\infty}}\frac{{2}^{k}{x}^{2k}}{1+{x}^{2k}}=x\sum _{k=0}^{\mathrm{\infty}}\frac{d}{dx}\left[\mathrm{log}(1+{x}^{2k})\right]$

$=x\frac{d}{dx}\left[\mathrm{log}(\prod _{k=0}^{\mathrm{\infty}}1+{x}^{{2}^{k}})\right]$

$=x\frac{d}{dx}\left[\mathrm{log}\left(\prod _{k=0}^{\mathrm{\infty}}\frac{1-{x}^{{2}^{k+1}}}{1-{x}^{{2}^{k}}}\right)\right]$

$=-x\frac{d}{dx}\left[\mathrm{log}(1-x)\right]$

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

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.

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.

usumbiix

Beginner2022-01-06Added 33 answers

Let $\left|x\right|<1$ and $k\ge 0$ . Using the factorization $1-{x}^{{2}^{k+1}}=(1-{x}^{{2}^{k}})(1+{x}^{{2}^{k}})$ , we find

$\frac{{2}^{k}{x}^{{2}^{k}}}{1+{x}^{{2}^{k}}}=\frac{{2}^{k}{x}^{{2}^{k}}[1+{x}^{{2}^{k}}-2{x}^{{2}^{k}}]}{(1+{x}^{{2}^{k}})(1-{x}^{{2}^{k}})}$

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

So since$\underset{k\to \mathrm{\infty}}{lim}\frac{{2}^{k}{x}^{{2}^{k}}}{1-{x}^{{2}^{k}}}=0$ , the series telescopes to $\frac{x}{1-x}$

So since

karton

Expert2022-01-11Added 613 answers

Let f(x) be the series. Rewriting, we get

Let

One can show that the above tends to 0 as

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