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

Vali

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Feb 27, 2018
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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

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Nov 12, 2017
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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

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Jan 29, 2005
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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

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Feb 27, 2018
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87
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...
tumblr_messaging_pmkn6x1DbL1tyt8xs_1280.jpg
 

pka

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Jan 29, 2005
Messages
8,717
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
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Feb 27, 2018
Messages
87
I think I did it.It's correct?
tumblr_messaging_pmks4ed8My1tyt8xs_1280.jpg
 

HallsofIvy

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Jan 27, 2012
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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 \(\displaystyle \frac{\pi^2}{6}- \frac{\pi^2}{24}= \frac{4\pi^2}{24}- \frac{\pi^2}{24}= \frac{3\pi^2}{24}= \frac{\pi^2}{8}\).
 
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