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 $\lim_{n \to \infty} nt_n$

Question

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?

Answer 1

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 $n t_n \to 1/8$.

Answer 2

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} $

Answer 3

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 $$\int_0^1\frac{dx}{(2+2x)^2}=\frac18.$$