Expert Answer to "How to prove that ∑k≥112kk2=π212−12log⁡(2)2"

Marquis Neal

Answered question

2022-01-23

How to prove that

\sum_{k \geq 1} \frac{1}{2^{k} k^{2}} = \frac{π^{2}}{12} - \frac{1}{2} {\log (2)}^{2}

immablondevl

Beginner2022-01-24Added 11 answers

Apply Euler Series Transformation to the series

\frac{π^{2}}{12} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{{(k + 1)}^{2}}

which yields

\frac{π^{2}}{12} = \sum_{n = 0}^{\infty} 2^{n - 1} \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) \frac{{(- 1)}^{k}}{{(k + 1)}^{2}}

= \sum_{n = 0}^{\infty} 2^{- n - 1} \frac{1}{n + 1} \sum_{k = 0}^{n} (\begin{matrix} n + 1 \\ k + 1 \end{matrix}) \frac{{(- 1)}^{k}}{k + 1}

= \sum_{n = 0}^{\infty} 2^{- n - 1} \frac{1}{n + 1} \sum_{j = 0}^{n} \sum_{k = 0}^{n} (\begin{matrix} j + 1 \\ k + 1 \end{matrix}) \frac{{(- 1)}^{k}}{j + 1}

= \sum_{n = 0}^{\infty} 2^{- n - 1} \frac{1}{n + 1} \sum_{j = 0}^{n} \frac{1}{J + 1}

= \sum_{n = 1}^{\infty} 2^{- n} \frac{1}{n} \sum_{j = 1}^{n} \frac{1}{j}

= \sum_{n = 1}^{\infty} \frac{H_{n}}{n 2^{n}}

Use the series for

\log (2) = - \log (1 - \frac{1}{2})

\log (2) = \sum_{m = 1}^{\infty} \frac{1}{k 2^{k}}

and square to get

{\log (2)}^{2} = \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{1}{m n 2^{m + n}}

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