TITLE: Computing $\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$
QUESTION [23 upvotes]: 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.

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\lim_{n\ \to\ \infty}\bracks{n\sum_{k = 1}^{n}\fermi\pars{k}}:\ {\large ?}.
     \qquad\fermi\pars{k} \equiv {1 \over \pars{2k -1}^{2}} - {3 \over 4k^{2}}}$

Note that $\ds{\sum_{k = 1}^{\infty}\fermi\pars{k} = 0}$ because
  $$
\sum_{k = 1}^{\infty}\fermi\pars{k}
=\sum_{k = 1}^{\infty}{1 \over \pars{2k - 1}^{2}}
-\sum_{k = 1}^{\infty}{3 \over 4k^{2}}
=\bracks{\sum_{k = 1}^{\infty}{1 \over k^{2}}
-\sum_{k = 1}^{\infty}{1 \over \pars{2k}^{2}}}
-\sum_{k = 1}^{\infty}{3 \over 4k^{2}}
$$

With Stolz-Cesàro Theorem:
\begin{align}&\color{#66f}{\large\lim_{n\ \to\ \infty}\bracks{%
n\sum_{k = 1}^{n} \fermi\pars{k}}}=\lim_{n\ \to\ \infty}{\sum_{k = 1}^{n}%
\fermi\pars{k} \over 1/n}
=\lim_{n\ \to\ \infty}{%
\sum_{k = 1}^{n + 1}\fermi\pars{k} - \sum_{k = 1}^{n}\fermi\pars{k}
\over 1/\pars{n + 1} - 1/n}
\\[3mm]&=\lim_{n\ \to\ \infty}\bracks{-n\pars{n + 1}\fermi\pars{n + 1}}
=\lim_{n\ \to\ \infty}\braces{n\pars{n + 1}
\bracks{{3 \over 4\pars{n + 1}^{2}} - {1 \over 4\pars{n + 1}^{2}}}}
\\[3mm]&=\half\,\lim_{n\ \to\ \infty}{n \over n + 1}=\color{#66f}{\Large\half}
\end{align}