Answered: "How to prove:∑n=1∞116n2−16n+3=π8"

dristauwaz

Answered question

2022-02-24

How to prove:

\sum_{n = 1}^{\infty} \frac{1}{16 n^{2} - 16 n + 3} = \frac{π}{8}

Halle Hansen

Beginner2022-02-25Added 5 answers

\sum_{n = 1}^{\infty} \frac{1}{16 n^{2} - 16 n + 3} = \frac{1}{8} \sum_{n = 1}^{\infty} (\frac{1}{n - \frac{3}{4}} - \frac{1}{n - \frac{1}{4}})

= \frac{1}{8} \sum_{n = 1}^{\infty} \int_{0}^{1} (x^{n - \frac{7}{4}} - x^{n - \frac{5}{4}}) d x

= \frac{1}{8} \int_{0}^{1} \sum_{n = 1}^{\infty} (x^{n - \frac{7}{4}} - x^{n - \frac{5}{4}}) d x

= \frac{1}{8} \int_{0}^{1} (\frac{x^{- \frac{3}{4}}}{1 - x} - \frac{x^{- \frac{1}{4}}}{1 - x}) d x

= \frac{1}{8} \int_{0}^{1} (\frac{x^{- \frac{3}{4}} - x^{- \frac{1}{4}}}{1 - x}) d x

= \frac{1}{8} \int_{0}^{1} \frac{1}{(\sqrt{x} + 1) x^{\frac{3}{4}}} d x

= \frac{1}{8} {[4 \arctan (\sqrt[4]{x})]}_{0}^{1}

= \frac{π}{8}

\sum_{n = 1}^{\infty} \frac{1}{16 n^{2} - 16 n + 3} = \frac{π}{8}

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