For each positive integer $n$, let $x_n=1/(n+1)+1/(n+2)+\cdots+1/(2n)$. Prove that the sequence $(x_n)$ converges.

This is what I have so far..

Let $\epsilon > 0$ be given to us. We must show that there exists an $N \in \mathbb{N}$ such that $n\ge N \implies \left|\frac{1}{2n} - 1\right| < \epsilon$. Choose $N$ to be any positive integer which is larger than $\frac{1}{\epsilon}$. (So $N>\frac{1}{\epsilon}$.) Then $n\ge N \implies \left|\frac{1}{2n} - 1\right| = |\frac{1-2n}{2n}|$

  • $\begingroup$ I need to find a way such to show that |1/(2n) – 1| < ε $\endgroup$ – Arnold Oct 14 '13 at 11:34


$$\frac12=\frac n{2n}\le\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}\le\frac n{n+1}$$

$$X_{n+1}:=\frac1{n+2}+\frac1{n+3}+\ldots+\frac1{2(n+1)}\le \frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}=:X_n$$

  • $\begingroup$ Can we find out to what number does the sequence converge? I know it is a monotonically increasing sequence bounded above by ln(2). Is ln(2) the supremum? Then, the limit would be ln(2). Is it? $\endgroup$ – Swapnil Tripathi Jun 9 '14 at 19:47
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    $\begingroup$ The last inequality is the wrong way around. Note that the number of terms on both sides is not equal, which is clouded by the suggestive notation. $\endgroup$ – WimC Feb 6 '16 at 18:00

We write $$x_n=\sum_{k=1}^n\frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+k/n}\to\int_0^1\frac{dx}{1+x}=\log 2$$


The sequence is decreasing because $$x_{n+1}-x_n = \frac1{2n+2}+\frac1{2n+1} - \frac1{n+1} = \frac{1}{2(2n+1)(n+1)}\ge 0$$ and obviously, $x_n\le \frac{n}{n+1}\le 1$, so is convergent. For proving that $x_n\to \log 2$ the better idea is bounding $x_n$ by definite integrals of the form $\int_a^b\frac1x\,dx$ (how?).

  • $\begingroup$ $-\frac{1}{n}$ is wrong, it should be $-\frac{1}{n+1}$, shouldn't it? But if so, you end up with $x_{n+1} \gt x_n$, right? Does it converge then? $\endgroup$ – Math for fun Oct 26 '17 at 21:14
  • $\begingroup$ @Flavius, corrected. Thanks. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 27 '17 at 6:30

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