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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 $n\ge N \implies \left|\frac{1}{2n} - 1\right| = |\frac{1-2n}{2n}|$

Arnold
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3 Answers3

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

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Hints:

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

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

[Edit: Flipping the inequality the right way]

Phalaksha C G
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DonAntonio
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  • 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? – Swapnil Tripathi Jun 09 '14 at 19:47
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    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. – WimC Feb 06 '16 at 18:00
  • @DonAntonio Nice answer! But you haven't computed the limit. Btw, do you know how to calculate it without using Riemann sums ? – Arthur Mar 10 '23 at 14:32
  • @Franklin This is very old...and the short answer is no: I don't know how to evaluate the limit without using Riemann sums. To prove the limit exists is done above in hints (bounded monotonic sequence), but the actual value of the limit...I personally can do it only with Riemann integrals. – DonAntonio Mar 10 '23 at 18:14
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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?).

Martín-Blas Pérez Pinilla
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