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?
 A: Another approach
$$t_n=\sum_{k=1}^n\left(\frac{1}{2n+2k-1}-\frac1{2n+2k}\right)=\sum_{k=1}^n\frac1{(2n+2k-1)(2n+2k)}$$
and then
$$nt_n=\frac1{n}\sum_{k=1}^n\frac1{(2+(2k-1)/n)(2+2k/n)}$$
which is a Riemann sum for
$$\int_0^1\frac{dx}{(2+2x)^2}=\frac18.$$
A: 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 $n t_n \to 1/8$.
A: Here is a
completely elementary proof 
by algebraic manipulation that
$\dfrac{1}{8n}-\dfrac{1}{16n^2}
\lt t_n
\lt \dfrac1{8n}
$.
$\begin{array}\\
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}\\
&=\sum_{k=2n+1}^{4n} \dfrac{(-1)^{k+1}}{k}\\
&=\sum_{k=n}^{2n-1} (\dfrac1{2k+1}-\dfrac1{2k+2})\\
&=\sum_{k=n}^{2n-1} \dfrac{1}{(2k+1)(2k+2)}\\
&<\sum_{k=n}^{2n-1} \dfrac{1}{(2k)(2k+2)}\\
&=\dfrac14\sum_{k=n}^{2n-1} \dfrac{1}{k(k+1)}\\
&=\dfrac14\sum_{k=n}^{2n-1} (\dfrac1{k}-\dfrac1{k+1})\\
&=\dfrac14(\dfrac1{n}-\dfrac1{2n})\\
&=\dfrac1{8n}\\
\text{and}\\
t_n
&=\sum_{k=n}^{2n-1} \dfrac{1}{(2k+1)(2k+2)}\\
&=\dfrac14\sum_{k=n}^{2n-1} \dfrac{1}{(k+1/2)(k+1)}\\
&>\dfrac14\sum_{k=n}^{2n-1} \dfrac{1}{(k+1/2)(k+3/2)}\\
&=\dfrac14\sum_{k=n}^{2n-1} (\dfrac1{k+1/2}-\dfrac1{k+3/2})\\
&=\dfrac14(\dfrac1{n+1/2}-\dfrac1{2n-1/2})\\
&=\dfrac14\dfrac{(2n-1/2)-(n+1/2)}{(n+1/2)(2n-1/2)}\\
&=\dfrac14\dfrac{n-1}{(n+1/2)(2n-1/2)}\\
&=\dfrac18\dfrac{n-1}{(n+1/2)(n-1/4)}\\
&=\dfrac18\dfrac{n+1/2-1/2}{(n+1/2)(n-1/4)}\\
&=\dfrac18(\dfrac{n+1/2}{(n+1/2)(n-1/4)}-\dfrac{1/2}{(n+1/2)(n-1/4)})\\
&=\dfrac18(\dfrac{1}{n-1/4}-\dfrac{1}{2(n+1/2)(n-1/4)})\\
&>\dfrac18(\dfrac{1}{n}-\dfrac{1}{2(n^2+n/4-1/8)})\\
&>\dfrac{1}{8n}-\dfrac{1}{16n^2}\\
\end{array}
$
