Let the real sequence ${x_n}$ be given by,

$$\sum_{j=1}^{2n} \frac {1}{j} - \sum_{j=1}^{n} \frac {1}{j}. $$

Show that $0<x_{n}<x_{n+1}$ and that $x_{n}<1$ for all $n$. Deduce that $x_{n}$ converges, giving your reason.

I seem to think this has something to do with $\sum_{j=1}^{2n} \frac {1}{j} - \sum_{j=1}^{n} \frac {1}{j} $ = $\sum_{j=1}^{n} \frac {1}{j+n}$?

Many thanks in advance.

  • 1
    $\begingroup$ This question is about the same sequence. $\endgroup$ Apr 10, 2012 at 15:45
  • $\begingroup$ The word "series" in the title doesn't fit: this is about convergence of a sequence, not of a series. $\endgroup$ Apr 10, 2012 at 16:23

2 Answers 2


First note that the sequence $(x_n)$ is bounded above. This follows from your observation that $\sum_{j=1}^{n} \frac {1}{j+n}$. Here we have $n$ terms, all of them clearly less than $1/n$, so their sum is less than $1$.

Next you want to show that the sequence $(x_n)$ is increasing. Calculate $x_{n+1}-x_n$, and show it is positive. Most of the terms cancel: $$x_{n+1}-x_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}.$$ Finally, appeal to the theorem that an increasing sequence which is bounded above has a limit.

Remark: The limit is in fact $\ln(2)$, but it seems you are not asked to show that. If you wish, it can be done by a Riemann sum argument.

  • $\begingroup$ Yes the next part is the Riemann sum this was fine however, many thanks for your quick reply. $\endgroup$
    – user24930
    Apr 10, 2012 at 15:58

$\sum^n 1/j \approx log(n)$ and that's all you need.


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