I have a question about this.
Using only the fact that
$$ 1 + \frac{1}{3^2} + {1 \over 5^2} + .... = {\pi^2 \over 8} ,$$
can we show $$ 1 + \frac{1}{2^2} + {1 \over 3^2} + {1 \over 4^2} + .... = { \pi^2 \over 6}. $$ I know that the second sum has to equal that, but I don't know how to show it using just that fact? thank you!
Yeah! Think about the latter sum as:
$$(1+1/3^2+1/5^2\cdots)+(1/2^2+1/4^2+1/6^2)=k$$
$\therefore (\pi^2/8)+(1/4)(1+1/4+1/9\cdots)=k$ ... and you notice something astonishing!
$$\pi^2/8+k/4=k$$ which Ron Gordon essentially said. Take it from here to solve $k$