For Each Positive Integer N, Let Xn=1/(n+1)+1/(n+2)+⋯+1/(2n). Prove ...

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Learn more about Teams For each positive integer $n$, let $x_n=1/(n+1)+1/(n+2)+\cdots+1/(2n)$. Prove that the sequence $(x_n)$ converges. Ask Question Asked 11 years, 1 month ago Modified 1 year, 4 months ago Viewed 27k times 4 $\begingroup$

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

• real-analysis
• sequences-and-series

Share Cite Follow edited Jan 26, 2018 at 17:39 user88319 asked Oct 14, 2013 at 11:28 ArnoldArnold 83966 gold badges1616 silver badges3030 bronze badges $\endgroup$ 1

• $\begingroup$ I need to find a way such to show that |1/(2n) – 1| < ε $\endgroup$ – Arnold Commented Oct 14, 2013 at 11:34

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Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 15 $\begingroup$

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

Share Cite Follow answered Oct 14, 2013 at 11:33 user63181user63181 $\endgroup$ 2

• 1 $\begingroup$ Big question, why is the integral up to 1? $\endgroup$ – Lilian Hernández Commented Apr 14, 2020 at 5:56
• $\begingroup$ @user63181 after putting x=k/n and dx/dk=1/n i.e dx=dk/n. but in the above integration you used simply dx. How ?? $\endgroup$ – The Iotaa Commented Sep 8, 2021 at 8:45

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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]

Share Cite Follow edited Jul 28, 2023 at 13:44 Ph C G 411010 bronze badges answered Oct 14, 2013 at 11:35 DonAntonioDonAntonio 214k1818 gold badges140140 silver badges290290 bronze badges $\endgroup$ 4

• $\begingroup$ 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 Commented Jun 9, 2014 at 19:47
• 2 $\begingroup$ 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 Commented Feb 6, 2016 at 18:00
• $\begingroup$ @DonAntonio Nice answer! But you haven't computed the limit. Btw, do you know how to calculate it without using Riemann sums ? $\endgroup$ – Arthur Commented Mar 10, 2023 at 14:32
• $\begingroup$ @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. $\endgroup$ – DonAntonio Commented Mar 10, 2023 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?).

Share Cite Follow edited Oct 27, 2017 at 6:30 answered Feb 9, 2016 at 13:56 Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla 42.2k44 gold badges4949 silver badges9292 bronze badges $\endgroup$ 2

• $\begingroup$ $-\frac{1}{n}$ 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 Commented Oct 26, 2017 at 21:14
• $\begingroup$ @Flavius, corrected. Thanks. $\endgroup$ – Martín-Blas Pérez Pinilla Commented Oct 27, 2017 at 6:30

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