# limit of a sum: 1 + 1/3 + ... + 1/(2n - 1)^2

#### Vali

##### Junior Member
I have the following sequence $$\displaystyle (x_{n})$$ , $$\displaystyle x_{n}=1+\frac{1}{2^{2}}+...+\frac{1}{n^{2}}$$ which has the limit $$\displaystyle \frac{\pi ^{2}}{6}$$.I need to calculate the limit of the sequence $$\displaystyle (y_{n})$$ , $$\displaystyle y_{n}=1+\frac{1}{3^{2}}+...+\frac{1}{(2n-1)^{2}}$$
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?

#### Dr.Peterson

##### Elite Member
I have the following sequence $$\displaystyle (x_{n})$$ , $$\displaystyle x_{n}=1+\frac{1}{2^{2}}+...+\frac{1}{n^{2}}$$ which has the limit $$\displaystyle \frac{\pi ^{2}}{6}$$.I need to calculate the limit of the sequence $$\displaystyle (y_{n})$$ , $$\displaystyle y_{n}=1+\frac{1}{3^{2}}+...+\frac{1}{(2n-1)^{2}}$$
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?
You want to subtract the sum of the even terms (zn, perhaps) from the sum of all terms (xn), to get the sum of odd terms (yn). Write out what the even terms are, and see how zn is related to xn.

#### pka

##### Elite Member
I have the following sequence $$\displaystyle (x_{n})$$ , $$\displaystyle x_{n}=1+\frac{1}{2^{2}}+...+\frac{1}{n^{2}}$$ which has the limit $$\displaystyle \frac{\pi ^{2}}{6}$$.I need to calculate the limit of the sequence $$\displaystyle (y_{n})$$ , $$\displaystyle y_{n}=1+\frac{1}{3^{2}}+...+\frac{1}{(2n-1)^{2}}$$
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?
Do you know about absolutely convergent series?

#### Vali

##### Junior Member
I'm very frustrated because I don't know how to solve this exercise.I tried every method in all ways possible, I miss something...

#### pka

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

#### Vali

##### Junior Member
I think I did it.It's correct?

#### HallsofIvy

##### Elite Member
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 $$\displaystyle \frac{\pi^2}{6}- \frac{\pi^2}{24}= \frac{4\pi^2}{24}- \frac{\pi^2}{24}= \frac{3\pi^2}{24}= \frac{\pi^2}{8}$$.

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