- 22,170
- 3,335
Ah, yes. this is good.
Now, you've got that
[tex]\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\sum_{n=1}^{+\infty}{\frac{1}{(2n)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(2n+1)^2}}[/tex]
You know two of the above series...
Now, you've got that
[tex]\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\sum_{n=1}^{+\infty}{\frac{1}{(2n)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(2n+1)^2}}[/tex]
You know two of the above series...