- #1

Vali

- 48

- 0

I don't know how to start.I think I need to solve the limit for all the sequence ( even n + odd n) then from the "big limit" I should subtract $\frac{\pi ^{2}}{6}$.How to start?

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- MHB
- Thread starter Vali
- Start date

- #1

Vali

- 48

- 0

I don't know how to start.I think I need to solve the limit for all the sequence ( even n + odd n) then from the "big limit" I should subtract $\frac{\pi ^{2}}{6}$.How to start?

- #2

skeeter

- 1,104

- 1

I don't know how to start.I think I need to solve the limit for all the sequence ( even n + odd n) then from the "big limit" I should subtract $\frac{\pi ^{2}}{6}$.How to start?

$x_n$ and $y_n$ are series ... sums of a sequence.

$\left(1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + ... + \dfrac{1}{n^2} + ... \right)-\left(1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + ... + \dfrac{1}{(2n-1)^2} + ... \right) = \dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + ... + \dfrac{1}{(2n)^2} + \, ...$

$\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^\infty \dfrac{1}{(2n-1)^2} = \sum_{n=1}^\infty \dfrac{1}{(2n)^2} = \dfrac{1}{4} \sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{24}$

$\displaystyle \text{note} \implies \sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^\infty \dfrac{1}{(2n)^2} = \sum_{n=1}^\infty \dfrac{1}{(2n-1)^2}$

- #3

Vali

- 48

- 0

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