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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.

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marked as duplicate by user296602, Clement C., Daniel W. Farlow, JMP, Claude Leibovici Aug 5 '16 at 5:35

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  • $\begingroup$ Did you try squeezing $x_n$ between two integrals? $\endgroup$ – Tim kinsella Aug 5 '16 at 2:40
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Note that we have

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

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HINT:

Note that

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

which is a Riemann sum.

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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.

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  • $\begingroup$ Necessarily the expression $ x_n > 0 $? $\endgroup$ – Jackson Aug 5 '16 at 2:54
  • $\begingroup$ 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. $\endgroup$ – Mr. Chip Aug 5 '16 at 2:55
  • $\begingroup$ Thanks, now I can continue. $\endgroup$ – Jackson Aug 5 '16 at 3:05

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