Don't use CAS on this one

Algebra Level 4

$\displaystyle \sum _{ n=1 }^{ \infty }{ \left( \sum _{ p=1 }^{ \infty }{ \left( \frac { 1 }{ { n }^{ 2p } } -\frac { 1 }{ { n }^{ 2p+1 } } \right) } \right) }= \ ?$

This problem is original.

\frac { 1 }{ 2 }

\frac { 1 }{ 3 }

2 1

\frac { 1 }{ 4 }

1 solution

Pop Wong
May 18, 2020

$\displaystyle \sum_{n=1}^\infty ( \displaystyle \sum_{p=1}^\infty ( \frac{1}{n^{2p}} - \frac{1}{n^{2p+1}} ) )$ $= \displaystyle \sum_{n=1}^\infty ( \displaystyle \sum_{p=1}^\infty ( \frac{n-1}{n^{2p+1}} ) )$ $= 0 + \displaystyle \sum_{n=2}^\infty ( (n-1) \displaystyle \sum_{p=1}^\infty ( \frac{1}{n^{2p+1}} ) )$

$= \displaystyle \sum_{n=2}^\infty ( (n-1) ( \frac{1}{n^3} + \frac{1}{n^5} +\ldots ) )$

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

$= \displaystyle \sum_{n=2}^\infty \frac{n-1}{n^3} \frac{n^2}{n^2-1}$

$= \displaystyle \sum_{n=2}^\infty \frac{1}{n (n+1)}$

$= \displaystyle \sum_{n=2}^\infty \frac{1}{n } - \frac{1}{n+1 }$

$= \frac{1}{2}$

Don't use CAS on this one

This problem is original.

1 solution

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