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 \(\displaystyle \frac{\pi ^{2}}{6}\).How to start?

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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 \(\displaystyle \frac{\pi ^{2}}{6}\).How to start?

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You want to subtract the sum of the even terms (z

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 \(\displaystyle \frac{\pi ^{2}}{6}\).How to start?

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Do you know about absolutely convergent series?

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 \(\displaystyle \frac{\pi ^{2}}{6}\).How to start?

Read about rearrangement properties.

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Did you read, study, and inwardly digest the link I posted? LOOK AT THISI'm very frustrated because I don't know how to solve this exercise.I tried every method in all ways possible, I miss something...

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Looks good to me. Good job!I think I did it.It's correct?

View attachment 10999

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In case people have difficulty reading that attachment (I did!):

We are given that the sum of reciprocals of squares of the integers, \(\displaystyle 1+ \frac{1}{2^2}+ \cdot\cdot\cdot+ \frac{1}{n^2}\) is equal to \(\displaystyle \frac{\pi^2}{6}\) and are asked to find the sum of reciprocals of the squares of**odd** integers, \(\displaystyle 1+ \frac{1}{3^2}+ \cdot\cdot\cdot+ \frac{1}{(2n-1)^2}\).

The idea is to subtract off the sum of reciprocal squares of**even** integers, \(\displaystyle \frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{6^2}+ \cdot\cdot\cdot+ \frac{1}{(2n)^2}\). But since an even integer is just 2 times an integer, we can factor out \(\displaystyle \frac{1}{2^2}= \frac{1}{4}\): \(\displaystyle \frac{1}{4}\left(1+ \frac{1}{2^2}+ \frac{1}{3^2}+ \cdot\cdot\cdot+ \frac{1}{n^2}\right)= \frac{1}{4}\frac{\pi^2}{6}= \frac{\pi^2}{24}\). So the sums of the reciprocals of the squares of the odd integers is the sum of the reciprocals of the squares of **all** integers, \(\displaystyle \frac{\pi^2}{6}\), minus the sum of the reciprocals of the squares of the **even** integers, \(\displaystyle \frac{\pi^2}{24}\), so is \(\displaystyle \frac{\pi^2}{6}- \frac{\pi^2}{24}= \frac{4\pi^2}{24}- \frac{\pi^2}{24}= \frac{3\pi^2}{24}= \frac{\pi^2}{8}\).

We are given that the sum of reciprocals of squares of the integers, \(\displaystyle 1+ \frac{1}{2^2}+ \cdot\cdot\cdot+ \frac{1}{n^2}\) is equal to \(\displaystyle \frac{\pi^2}{6}\) and are asked to find the sum of reciprocals of the squares of

The idea is to subtract off the sum of reciprocal squares of

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