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.
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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 \(\displaystyle{\frac{{{x}}}{{{1}-{x}}}}\)
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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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