Infinite Sum Calculation: $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$ Problem
Show the following equivalence: $$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$

Good afternoon, dear StackExchange community. I'm studying real analysis (the topic right now is exchanging limits) and I can't wrap my head around this problem. It might be a duplicate, but I couldn't find a thread that helped me to solve this problem.
Also, I know that both terms equals $\frac{\pi^2}{8}$ (courtesy of Wolframalpha), but I think I should show the original problems via exchanging limits of double series or with analyising the summands, because we didn't introduce the identity $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi}{6}$ yet.
My Attempts
If we consider the first summands of both sums, we have:
\begin{array}{|c|c|c|c|}
\hline
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{(2k+1)^2} & 1 & \frac{1}{9} & \frac{1}{25} & \frac{1}{49} & \frac{1}{81} & \frac{1}{121} & \frac{1}{169} \\ \hline
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{n^2} & / & 1 & \frac{1}{4} & \frac{1}{9} & \frac{1}{16} & \frac{1}{25} & \frac{1}{36} \\ \hline
\end{array}
We immediately see that in $\frac{1}{n^2}$ is every summand of $\frac{1}{(2k+1)^2}$ is included. Additionally, we see that the extra summands in $\frac{1}{n^2}$ have the form $\frac{1}{(2x)^2}$.
Therefore, we could write,
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$$
and the original equation would be
$$\Rightarrow 1 + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \left( 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\right)$$.
But I don't know how this could help me.

Anyways, series and infinite sums give me headaches and I would appreciate any help or hints. Thank you in advance.

Wow! Thank you for the really quick replies!
 A: Whoa, back up a bit!
$$\begin{align}\sum_{n=1}^{\infty}\frac{1}{n^2} &= 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\\&= \frac1{2(0)+1} + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\\&=\sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}\end{align}$$
Notice that $\frac1{(2k)^2}=\frac1{4k}$ so that
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac14\sum_{k=1}^{\infty} \frac{1}{k^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
Subtract $\frac14\sum_{k=1}^{\infty} \frac{1}{k^2}$ from both sides to get

$$\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=\frac34\sum_{k=1}^{\infty}\frac{1}{k^2}$$

A: You are almost done,:
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}= \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
Notice that :
$$\sum_{k=1}^{\infty} \frac{1}{(2k)^2}=\sum_{k=1}^{\infty} \frac{1}{4k^2}=\frac14\sum_{k=1}^{\infty} \frac{1}{k^2}$$
Finally :
$$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
A: $\displaystyle \begin{align}\sum_{n\ge 0} \dfrac{1}{(2n+1)^s}&= \sum_{n=1}^\infty \dfrac{1}{n^s}-\sum_{n=1}^\infty \dfrac{1}{(2n)^s} \\ &= \sum_{n=1}^\infty \dfrac{1}{n^s}(1-2^{-s}) \\ &= (1-2^{-s})\sum_{n=1}^{\infty} \dfrac{1}{n^s}\end{align}$
Putting $s=2$ gives you the result and obviously the above is true for $s\in\mathbb{N}$
