Check the solution to "Please help me, to prove that∑n=2∞2(n3−n)3n=−12+43∑n=1∞1n⋅3n"

Kris Patton

Answered question

2022-02-24

Please help me, to prove that

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

junoon363km

Beginner2022-02-25Added 8 answers

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

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

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

= \sum_{n = 1}^{\infty} \frac{1}{n 3^{n}} (\frac{1}{3} - 2 + 3) + (\frac{2}{3} - 1 - \frac{1}{6})

= \frac{4}{3} \sum_{n = 1}^{\infty} \frac{1}{n \cdot 3^{n}} - \frac{1}{2}

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