solve the limit of $(y_{n})$, $y_{n}=1+\frac{1}{3^{2}}+...+\frac{1}{(2n-1)^{2}}$

Question

I have the following sequence $(x_{n})$ , $x_{n}=1+\frac{1}{2^{2}}+...+\frac{1}{n^{2}}$ which has the limit $\frac{\pi ^{2}}{6}$.I need to calculate the limit of the sequence $(y_{n})$, $y_{n}=1+\frac{1}{3^{2}}+...+\frac{1}{(2n-1)^{2}}$

I don't know how to start.I think I need to solve the limit for the whole sequence ( even n + odd n) then from the "big limit" I should subtract $\frac{\pi ^{2}}{6}$, right?

Answer 1

Use $$\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{(2n)^2}=\frac{1}{4}\left(1+\frac{1}{2^2}+...+\frac{1}{n^2}\right).$$ Also, we can write the following. $$1+\frac{1}{3^2}+...+\frac{1}{(2n-1)^2}=$$ $$=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{(2n-1)^2}+\frac{1}{(2n)^2}-\left(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{(2n)^2}\right)=$$ $$=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{(2n-1)^2}+\frac{1}{(2n)^2}-\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}\right)\rightarrow...$$ I got $\frac{\pi^2}{8}.$