Computing $\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$ What ways would you propose for the limit below?
$$\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$$
Thanks in advance for your suggestions, hints! 
Sis.

Using the Euler-Maclaurin Sum Formula,
$$
\begin{align}
\sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right)
&=\sum_{k=1}^n\left(\frac1{(2k-1)^2}+\frac1{(2k)^2}-\frac1{k^2}\right)\\
&=\sum_{k=1}^{2n}\frac1{k^2}-\sum_{k=1}^n\frac1{k^2}\\
&=\left(C-\frac1{2n}+O\left(\frac1{n^2}\right)\right)-\left(C-\frac1n+O\left(\frac1{n^2}\right)\right)\\
&=\frac1{2n}+O\left(\frac1{n^2}\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\lim_{n\to\infty}n\sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right)
&=\lim_{n\to\infty}n\left(\frac1{2n}+O\left(\frac1{n^2}\right)\right)\\
&=\frac12
\end{align}
$$
