I have this homework problem, that I'm stuck on.

We know that:$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

I have to find the sum of: $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots$$

I came up with this equation: $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}\right)^2$$

I know that the answer is $\frac{\pi^2}{8}$ I found $a=1$, but can't seem to figure it out...




  • 1
    $\begingroup$ How do you know that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\sum_{n=1}^{\infty}\frac{1}{(2n)^2}$? $\endgroup$ – KFC Nov 15 '14 at 18:47
  • 2
    $\begingroup$ Just split the sum in odd and even terms. $\endgroup$ – ajotatxe Nov 15 '14 at 18:50

Of more historical than mathematical interest, Euler's original approach:

$\cos x = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+...$

So $\cos x$ may be regarded as a polynomial with roots $\pm \pi/2, \pm 3\pi/2, \pm 5\pi/2,...$

That is,

$$\cos x = (x - \frac{\pi}{2})(x + \frac{\pi}{2})(x - \frac{3\pi}{2})(x + \frac{3\pi}{2})...$$

$$= (x^2-\frac{\pi^2}{4})(x^2-\frac{3^2\pi^2}{4})(x^2-\frac{5^2\pi^2}{4})... $$

which we can write as

$$(*) \hspace{10mm}1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+... = A(1-\frac{4x^2}{\pi^2})(1-\frac{4x^2}{3^2\pi^2})(1-\frac{4x^2}{5^2\pi^2})...$$

for some constant A; and since $\cos x \to 1 ~~\text{as}~~ x\to 0,$ A must equal $1$.

Then from (*) we can equate coefficients of $x^2:$

$$-\frac{1}{2!} = -\frac{4}{\pi^2}- \frac{4}{3^2\pi^2}- \frac{4}{5^2\pi^2}-... $$

so finally

$$\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}+... $$


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