# Consider the following infinite series: \sum_{k=0}^\infty\frac{2^k}{1+\frac{1}{x^{2k}}}

Joan Thompson 2022-01-04 Answered
Consider the following infinite series:
$$\displaystyle{\sum_{{{k}={0}}}^{\infty}}{\frac{{{2}^{{k}}}}{{{1}+{\frac{{{1}}}{{{x}^{{{2}{k}}}}}}}}}$$

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## Expert Answer

Hattie Schaeffer
Answered 2022-01-05 Author has 3091 answers
To me, the sum can be derived informally as
$$\displaystyle{\sum_{{{k}={0}}}^{\infty}}{\frac{{{2}^{{k}}{x}^{{{2}{k}}}}}{{{1}+{x}^{{{2}{k}}}}}}={x}{\sum_{{{k}={0}}}^{\infty}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\log{{\left({1}+{x}^{{{2}{k}}}\right)}}}\right]}$$
$$\displaystyle={x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\log{{\left({\prod_{{{k}={0}}}^{\infty}}{1}+{x}^{{{2}^{{k}}}}\right)}}}\right]}$$
$$\displaystyle={x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\log{{\left({\prod_{{{k}={0}}}^{\infty}}{\frac{{{1}-{x}^{{{2}^{{{k}+{1}}}}}}}{{{1}-{x}^{{{2}^{{k}}}}}}}\right)}}}\right]}$$
$$\displaystyle=-{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\log{{\left({1}-{x}\right)}}}\right]}$$
$$\displaystyle={\frac{{{x}}}{{{1}-{x}}}}$$
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.
###### Not exactly what you’re looking for?
usumbiix
Answered 2022-01-06 Author has 1667 answers
Let $$\displaystyle{\left|{x}\right|}{ < }{1}$$ and $$\displaystyle{k}\geq{0}$$. Using the factorization $$\displaystyle{1}-{x}^{{{2}^{{{k}+{1}}}}}={\left({1}-{x}^{{{2}^{{k}}}}\right)}{\left({1}+{x}^{{{2}^{{k}}}}\right)}$$, we find
$$\displaystyle{\frac{{{2}^{{k}}{x}^{{{2}^{{k}}}}}}{{{1}+{x}^{{{2}^{{k}}}}}}}={\frac{{{2}^{{k}}{x}^{{{2}^{{k}}}}{\left[{1}+{x}^{{{2}^{{k}}}}-{2}{x}^{{{2}^{{k}}}}\right]}}}{{{\left({1}+{x}^{{{2}^{{k}}}}\right)}{\left({1}-{x}^{{{2}^{{k}}}}\right)}}}}$$
$$\displaystyle={\frac{{{2}^{{k}}{x}^{{{2}^{{k}}}}}}{{{1}-{x}^{{{2}^{{k}}}}}}}-{\frac{{{2}^{{{k}+{1}}}{x}^{{{2}^{{{k}+{1}}}}}}}{{{1}-{x}^{{{2}^{{{k}+{1}}}}}}}}$$
So since $$\displaystyle\lim_{{{k}\to\infty}}{\frac{{{2}^{{k}}{x}^{{{2}^{{k}}}}}}{{{1}-{x}^{{{2}^{{k}}}}}}}={0}$$, the series telescopes to $$\displaystyle{\frac{{{x}}}{{{1}-{x}}}}$$
karton
Answered 2022-01-11 Author has 8659 answers

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

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