For $n$ a positive integer, put: $$t_n = \frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\ldots+\frac{1}{4n-1}-\frac{1}{4n}$$ Find, with proof, the following limit $\mathcal{T}$: $$\mathcal{T}=\lim\limits_{n\to\infty}nt_n.$$ Hint: Relate the given limit to suitable Riemann sums for the function $(1+x)^{-2}$.
My attempt is to rewrite $nt_n$ as $\sum\limits_{i=1}^{2n}(-1)^{i+1}\frac{n}{2n+i}$, then, since this is an alternating partial sums, split it into two sums as $$\sum\limits_{i=1}^{2n}(-1)^{i+1}\frac{n}{2n+i} = \sum\limits_{i=0}^{n-2}\frac{n}{2n+(2i+1)}-\sum\limits_{i=1}^n\frac{n}{2n+2i} = \frac{1}{2}\sum\limits_{i=0}^{n-2}\frac{1}{1+\frac{1}{2n}+\frac{i}{n}}-\frac{1}{2}\sum\limits_{i=1}^n\frac{1}{1+\frac{i}{n}},$$ however, I can't see how to link it with the Riemann sums for $\frac{1}{(1+x)^2}$.
How about adding together adjacent positive and negative terms? I get $$nt_n=\sum_{k=1}^n n\left(\frac1{2n+2k-1}-\frac1{2n+2k}\right) =\sum_{k=1}^n\frac n{(2n+2k-1)(2n+2k)} $$which looks more the sort of thing that could be written in terms of Riemann sums.