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.
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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}
A: Here is a high school level answer:
$$
\begin{align}
\sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right)
&=\sum_{k=1}^n\left(\frac1{(2k-1)^2}+\frac1{(2k)^2}-\frac1{k^2}\right)\\
&=\sum_{k=1}^{2n}\frac1{k^2}-\sum_{k=1}^n\frac1{k^2}\\
&=\sum_{k=n+1}^{2n}\frac1{k^2}\tag{1}
\end{align}
$$
Using partial fractions and summing the telescoping series, we get
$$
\hspace{-1cm}
\frac1{n+1}-\frac1{2n+2}
=\sum_{k=n+1}^{2n}\frac1{k(k+1)}
\le\sum_{k=n+1}^{2n}\frac1{k^2}
\le\sum_{k=n+1}^{2n}\frac1{k(k-1)}
=\frac1n-\frac1{2n}\tag{2}
$$
Therefore, the Squeeze Theorem and $(2)$ yield
$$
\lim_{n\to\infty}n\sum_{k=n+1}^{2n}\frac1{k^2}=\frac12\tag{3}
$$
A: OK, it turns out that
$$\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right) = \sum_{k=1}^{n-1} \frac{1}{(k+n)^2}$$
This may be shown by observing that
$$\sum_{k=1}^n \frac{1}{(2k-1)^2} = \sum_{k=1}^{2 n-1} \frac{1}{k^2} - \frac{1}{2^2} \sum_{k=1}^n \frac{1}{k^2}$$
The desired limit may then be rewritten as
$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{(1 + (k/n))^2}$$
which you may recognize as a Riemann sum, equal to 
$$\int_0^1 dx \: \frac{1}{(1+x)^2} = \frac{1}{2}$$
A: $$n\sum_{k=1}^n\frac{1}{(2k-1)^2}-\frac{3}{4k^2} =n(H_{2n}^{(2)}-H_{n}^{(2)})=\sum_{j=1}^n\frac{n}{(j+n)^2}\to\int_0^1\frac{dx}{(1+x)^2} =\frac{1}{2}$$
A: 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,
$$
\begin{align}
\lim_{n\to\infty}n\sum_{k=1}^n\left(\frac1{(2k-1)^2}-\frac3{4k^2}\right)
&=\lim_{n\to\infty}n\left(\frac1{2n}+O\left(\frac1{n^2}\right)\right)\\
&=\frac12
\end{align}
$$
