Sequence and Series bounded

Question

Problem: Let $ x_n $ be a sequence defined by $x_n = \displaystyle {\sum_{k=n+1}^{ 2n}\frac{1}{k}}$. Show that $ x_n$ converge.

I can bounded $ x_n <\frac{ n}{ n+1}$, but I can't show that $ x_n$ is increasing, thus would use that es increasing and bounded then is convergent.

Did you try squeezing $x_n$ between two integrals? – Tim kinsella Aug 05 '16 at 02:40 — Tim kinsella, Aug 05 '16 at 02:40

score 1 · Answer 1 · answered Aug 05 '16 at 02:40

1

HINT:

Note that

$$\sum_{n=k+1}^{2n}\frac1k =\frac1n \sum_{k=1}^n\frac{1}{1+k/n}$$

which is a Riemann sum.

answered Aug 05 '16 at 02:40

Mark Viola

179,405

Mr. Chip · Answer 2 · 2016-08-05T02:48:45.160

1

Simply notice that, since $\frac{1}{2n+1} > \frac{1}{2n+2}$, $$x_{n+1} - x_n = \sum_{k=n+2}^{2n+2} \frac{1}{k} - \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} > \frac{2}{2n+2} - \frac{1}{n+1} = 0.$$ Hence the sequence is increasing. Then apply your bound $x_n \le \frac{n}{n+1} < 1$ to conclude convergence.

edited Aug 05 '16 at 02:48

answered Aug 05 '16 at 02:42

Mr. Chip

5,009

Necessarily the expression $ x_n > 0 $? – Jackson Aug 05 '16 at 02:54
I have shown $x_{n+1} - x_n > 0$, which proves $x_{n+1} > x_n$, i.e. the sequence is increasing. It's also true that $x_n > 0$, since it's a sum of positive numbers. – Mr. Chip Aug 05 '16 at 02:55
Thanks, now I can continue. – Jackson Aug 05 '16 at 03:05

score 1 · Accepted Answer · answered Aug 05 '16 at 02:44

Note that we have

$$\begin{align} \sum_{k=n+1}^{2n}\frac1k &=\sum_{k=1}^{2n}\frac1k -\sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2k}\right)-\sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\ &=\sum_{k=1}^n \frac{(-1)^{n-1}}{k}\\\\ &\to \log(2) \end{align}$$

Sequence and Series bounded

3 Answers3