3
$\begingroup$

I ran into this question:

Prove that:

$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Thank you very much in advance.

$\endgroup$
16
+50
$\begingroup$

Hint: $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}$$

$\endgroup$
  • 10
    $\begingroup$ @user76508 No problem, I am loyal to the Workers. $\endgroup$ – Kortlek May 12 '13 at 22:23
  • $\begingroup$ @RobFord I thought you were libertarian? $\endgroup$ – Anonymous - a group Oct 9 '14 at 16:05
5
$\begingroup$

Hints:

$$n\in\Bbb N:=\{1,2,\ldots\}:\;\;\;\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\ldots$$

$\endgroup$
  • $\begingroup$ I simply not understanding how the right and left side are equal of the series. $\endgroup$ – Marso May 17 '13 at 5:06
  • $\begingroup$ please help me to understand. $\endgroup$ – Marso May 17 '13 at 5:07
  • 1
    $\begingroup$ @Tsotsi: just divide the sum of the reciprocals of the squared naturals in even ones $\,(2n)^{-2}\,$ and odd ones $\,(2n-1)^{-2}\,$ , but then $\,(2n)^{-2}=2^{-2}\cdot n^{-2}\,$ and etc. $\endgroup$ – DonAntonio May 17 '13 at 7:20
  • $\begingroup$ $1^2+{1\over 2^2}+{1\over 3^2}+\dots+{1\over n^2}+\dots={1\over (2n)^2}+{1\over (2n-1)^2}+\dots$ am I right? $\endgroup$ – Marso May 17 '13 at 8:00
  • $\begingroup$ @DonAntonio I should be hanged :( $\endgroup$ – Marso May 17 '13 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.