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.

Solution 1:

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 $$ \lim_{n\to\infty}n\sum_{k=n+1}^{2n}\frac1{k^2}=\frac12\tag{3} $$

Solution 2:

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

$$\int_0^1 dx \: \frac{1}{(1+x)^2} = \frac{1}{2}$$

Solution 3:

$$n\sum_{k=1}^n\frac{1}{(2k-1)^2}-\frac{3}{4k^2} =n(H_{2n}^{(2)}-H_{n}^{(2)})=\sum_{j=1}^n\frac{n}{(j+n)^2}\to\int_0^1\frac{dx}{(1+x)^2} =\frac{1}{2}$$