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

To me, the sum can be derived informally as
$\sum _{k=0}^{\mathrm{\infty }}\frac{{2}^{k}{x}^{2k}}{1+{x}^{2k}}=x\sum _{k=0}^{\mathrm{\infty }}\frac{d}{dx}\left[\mathrm{log}\left(1+{x}^{2k}\right)\right]$
$=x\frac{d}{dx}\left[\mathrm{log}\left(\prod _{k=0}^{\mathrm{\infty }}1+{x}^{{2}^{k}}\right)\right]$
$=x\frac{d}{dx}\left[\mathrm{log}\left(\prod _{k=0}^{\mathrm{\infty }}\frac{1-{x}^{{2}^{k+1}}}{1-{x}^{{2}^{k}}}\right)\right]$
$=-x\frac{d}{dx}\left[\mathrm{log}\left(1-x\right)\right]$
$=\frac{x}{1-x}$
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.

usumbiix

Let $|x|<1$ and $k\ge 0$. Using the factorization $1-{x}^{{2}^{k+1}}=\left(1-{x}^{{2}^{k}}\right)\left(1+{x}^{{2}^{k}}\right)$, we find
$\frac{{2}^{k}{x}^{{2}^{k}}}{1+{x}^{{2}^{k}}}=\frac{{2}^{k}{x}^{{2}^{k}}\left[1+{x}^{{2}^{k}}-2{x}^{{2}^{k}}\right]}{\left(1+{x}^{{2}^{k}}\right)\left(1-{x}^{{2}^{k}}\right)}$
$=\frac{{2}^{k}{x}^{{2}^{k}}}{1-{x}^{{2}^{k}}}-\frac{{2}^{k+1}{x}^{{2}^{k+1}}}{1-{x}^{{2}^{k+1}}}$
So since $\underset{k\to \mathrm{\infty }}{lim}\frac{{2}^{k}{x}^{{2}^{k}}}{1-{x}^{{2}^{k}}}=0$, the series telescopes to $\frac{x}{1-x}$

karton

Let f(x) be the series. Rewriting, we get
$f\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{2}^{k}{x}^{2k}}{1+{x}^{{2}^{k}}}$
Let ${s}_{n}$ be the n-th partial sum. We then have
$\frac{x}{1-x}-{s}_{n}=\frac{x}{1-x}-\frac{x}{1+x}-...-\frac{{2}^{n}{x}^{{2}^{n}}}{1+{x}^{{2}^{n}}}$
$=\frac{{2}^{n+1}{x}^{{2}^{{2}^{n+1}}}}{1+{x}^{{2}^{n+1}}}$
One can show that the above tends to 0 as $n\to \mathrm{\infty }$ for |x|<1

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