I'm working on a problem for a preliminary exam, and I came across one that confused me.
For $n\in\mathbb{N}$, set
$t_n=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{2n+i}=\frac{1}{2n+1}-\frac{1}{2n+2}+\cdots+\frac{1}{4n-1}-\frac{1}{4n}$
Evaluate $\lim_{n\to\infty} nt_n$
$\textit{Hint:}$ Relate the given limit to suitable Riemann sums for $f(x)=(1+x)^{-2}$.
I've evaluated limits using Riemann sums before, and I've simplified
$\lim_{n\to\infty}nt_n=\lim_{n\to\infty}n(\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{2n+i})=\lim_{n\to\infty}n\frac{1}{2n}\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{1+\frac{i}{2n}}=\lim_{n\to\infty}\frac{1}{2}\sum_{i=1}^{2n}(-1)^{i+1}\sqrt{f(\frac{i}{2n})}$
This is as far as I've been able to complete however. I am confused as to the alternating nature of the sum, and how to complete the limit problem. Any help would be appreciated.
Well, here's one way to do it,
$t_n=\sum\limits_{i=1}^{2n}\frac{\left(-1\right)^{i+1}}{2n+i}=\sum\limits_{i\leq2n,i\in\mathrm{Odds}}\frac1{2n+i}-\sum\limits_{i\leq2n,i\in\mathrm{Evens}}\frac1{2n+i}=\sum\limits_{i=1}^n\frac1{2n+2i-1}-\sum\limits_{i=1}^n\frac1{2n+2i}=\sum\limits_{i=1}^n\frac1{\left(2n+2i-1\right)\left(2n+2i\right)}=\frac1{4n^2}\sum\limits_{i=1}^n\frac1{\left(1+\frac{i-\frac12}n\right)\left(1+\frac in\right)}$
So,
$nt_n=\frac1{4n}\sum\limits_{i=1}^n\frac1{\left(1+\frac{i-\frac12}n\right)\left(1+\frac in\right)}$.
This is a Riemann sum, that is,
$\lim\limits_{n\rightarrow\infty}nt_n=\frac14\int_0^1\frac{\mathrm dx}{\left(1+x\right)^2}=\frac18$