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Octavio Miller

Answered question

2022-01-22

Evaluate

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

Answer & Explanation

enguinhispi

Beginner2022-01-23Added 15 answers

Given:
$S = \sum_{n = 1}^{\infty} \frac{1}{n^{2} 2^{n}}$
Consider the following Lemma,
Lemma:
$\int_{0}^{1} \frac{\ln x}{x + 1} d x = - \frac{π^{2}}{12}$
Proof:
$\int_{0}^{1} \frac{\ln x}{x + 1} d x = \int_{0}^{1} \frac{- \ln x}{1 - x} d x + \int_{0}^{1} \frac{- 2 \ln x}{x^{2} - 1} d x$
Also,
$\int_{0}^{1} \frac{- \ln x}{1 - x} d x$
$= \int_{0}^{1} \frac{- \ln (1 - x)}{x} d x = \int_{0}^{1} \frac{x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots}{x} d x$
$= \sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6}$
and,
$\int_{0}^{1} \frac{\ln x}{x^{2} - 1} d x = \sum_{n = 0}^{\infty} \int_{0}^{1} x^{2 n} \ln x d x$
$= - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ (Using Integration By Parts)
$= \sum_{n = 0}^{\infty} \frac{1}{{(2 n)}^{2}} - \sum_{n = 0}^{\infty} \frac{1}{n^{2}}$
$= \frac{- 3}{4} ζ (2) = - \frac{π^{2}}{8}$
$∴ \int_{0}^{1} \frac{\ln x}{x + 1} d x = \frac{π^{2}}{6} - \frac{π^{2}}{4} = - \frac{π^{2}}{12}$
This completes the proof of our Lemma.
Now,

Karli Kaiser

Beginner2022-01-24Added 9 answers

You series equals

L i_{2} (\frac{1}{2})

. By the functional identity:

L i_{2} (z) + L i_{2} (1 - z) = ζ (2) - \log (z) \log (1 - z)

that is straightforward to prove by differentiation, since:

\frac{d}{d z} L i_{2} (z) = \frac{d}{d z} \sum_{n \geq 1} \frac{z^{n}}{n^{2}} = - \frac{\log (1 - z)}{z}

we have:

\sum_{n \geq 1} \frac{1}{n^{2} 2^{n}} = L i_{2} (\frac{1}{2}) = \frac{1}{2} (\frac{π^{2}}{6} - \log^{2} 2)

RizerMix

Expert2022-01-27Added 656 answers

Recall that

\sum_{k = 1}^{\infty} \frac{x^{k}}{k} = - \log (1 - x)

(1) Dividing (1) by x and integrating yields

\sum_{k = 1}^{\infty} \frac{1}{2^{k} k^{2}} = - \int_{0}^{1 / 2} \frac{\log (1 - x)}{x} d x

= - [\log (1 - x) \log (x)]_{0}^{1 / 2} - \int_{0}^{1 / 2} \frac{\log (x)}{1 - x} d x

= - \log (2)^{2} - \int_{1 / 2}^{1} \frac{\log (1 - x)}{x} d x

= - \frac{\log (2)^{2}}{2} - \frac{1}{2} \int_{0}^{1} \frac{\log (1 - x)}{x} d x

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

= - \frac{\log (2)^{2}}{2} + \frac{π^{2}}{12}

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