If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two terms and performing subtraction we will get it) so is $nt_n$. Now can we prove it is monotonically decreasing? If so then $\lim_{n \to \infty} nt_n=\lim_{n \to \infty}[(\frac{1}{2+1/n}-\frac{1}{2+2/n})+(\frac{1}{2+3/n}-\frac{1}{2+4/n})+\cdots +(\frac{1}{4-1/n}-\frac{1}{4})]$ and each of these terms will go to zero so is the limit.
Second attempt: I was trying to use Riemann summation $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}=
\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{2n+3}+\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}+\frac{1}{4n}-2[\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots \frac{1}{4n}]\Rightarrow \lim \frac 1n [nt_n]=\int_0^2\frac{dx}{2+x}-\int_0^1\frac{dx}{1+x}=\ln4-\ln 2-\ln 2=0$
So $\lim t_n=0$
So what will happen with $\lim nt_n$
Edit: As I got the answer is not $0$ because of the flaw. So can we have different approaches even with Riemann Sum to have the answer?

Let $$H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}$$ be the $n$th harmonic number. Then, as you noted, $$t_n = H_{4n} - H_{2n} - [H_{2n} - H_n] = H_{4n} - 2H_{2n} + H_n.$$
By the Euler-Maclaurin summation formula, 
$$H_n = \log n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$ 
so, after the smoke clears, $$t_n = \frac{1}{8n} + O\left(\frac{1}{n^2}\right)$$ whence $n t_n \to 1/8$.
