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

[Vali]

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]

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 (z_n, perhaps) from the sum of all terms (x_n), to get the sum of odd terms (y_n). Write out what the even terms are, and see how z_n is related to x_n.

 

[pka]

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?
Read about rearrangement properties.

 

[Vali]

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]

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]

I think I did it.It's correct?

 

[Steven G]

I think I did it.It's correct?
View attachment 10999

Looks good to me. Good job!

 

[HallsofIvy]

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 \(\displaystyle \frac{\pi^2}{6}- \frac{\pi^2}{24}= \frac{4\pi^2}{24}- \frac{\pi^2}{24}= \frac{3\pi^2}{24}= \frac{\pi^2}{8}\).