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The natural way (and I think, the easier) to solve $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$$

involves Riemann sums. Multiplying and dividing by $n$, we get

$$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)=\frac{1}{n}\sum_{i=1}^{n}\frac{n}{n+i}=\frac{1}{n}\sum_{i=1}^nf\left(\frac{i}{n}\right)$$

where $f(x)=\frac{1}{1+x}$. The last expression is a Riemann sum associated to the partition $P_n=\{0, \frac{1}{n},\dots,\frac{n}{n}\}$ of $[0,1]$, so by Darboux Theorem

$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^nf\left(\frac{i}{n}\right)=\int_0^1\frac{1}{1+x}dx=\ln2$$

However, I'm curious if there are other techniques that can be used to find that limit, so my question is

Is it possible to find $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$$ without using Riemann sums? If so, how?

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Starting from $e^x\ge 1+x$ and thus also $e^{-x}\ge 1-x$ one gets $$ -\ln(1-x)\ge x \ge \ln(1+x) $$ Now set $x=\frac1{n+k}$ to get $$ \ln(n+k)-\ln(n+k-1)\ge\frac1{n+k}\ge\ln(n+k+1)-\ln(n+k) $$ Adding up for the required expression gives telescoping in sums on the left and right, so $$ \ln(2)=\ln(2n)-\ln(n)\ge\sum_{k=1}^n\frac1{n+k}\ge\ln(2n+1)-\ln(n+1)=\ln\left(2-\frac1{n+1}\right) $$

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  • $\begingroup$ I took the liberty of correcting a typo in the second line; I hope that you don't mind. - Mark $\endgroup$ – Mark Viola Feb 24 '16 at 21:00
  • $\begingroup$ Danke. I had the "+k" inserted in many places, but forgot that one. $\endgroup$ – LutzL Feb 25 '16 at 0:02
  • $\begingroup$ Nice! Simple and clever answer. Thank you :) $\endgroup$ – user281593 Feb 25 '16 at 16:38
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We have \begin{align} H_n &= 1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}\\\\ \Rightarrow \quad S_n &= H_{2n} - H_n \\\\ & = (H_{2n} - \ln(2n) - \gamma )- (H_n - \ln n - \gamma) + \ln 2\\\\ & = a_n + b_n + \ln 2\to 0+0+\ln 2 = \ln 2\end{align} as $n \to \infty$.

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    $\begingroup$ How do you get that $\lim_{n \to \infty} H_n - \log n - \gamma = 0$ without using Riemann sums? $\endgroup$ – Michael Lugo Feb 24 '16 at 18:37
  • $\begingroup$ @MichaelLugo The Euler-Maclaurin summation formula provides the expansion for the Harmonic number. $\endgroup$ – Mark Viola Feb 25 '16 at 17:53
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One way that does not use the harmonic series is by using the identity $$\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i},$$ which you can find here, for example, and use the power series representation of $\ln(1+x)$ at $x=1$.

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  • $\begingroup$ Since the radius of convergence of $\ln(1+x)$ is exactly $1$, there is the (small) technicality of invoking Abel's Theorem, isn't there? $\endgroup$ – Clement C. Feb 24 '16 at 18:18
  • $\begingroup$ Yes, you are right. $\endgroup$ – detnvvp Feb 24 '16 at 18:22
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Here is one approach. Write the sum of interest as

$$\begin{align} \sum_{k=n+1}^{2n}\frac1k&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac1{2k}\right)\\\\ &=\sum_{k=1}^n\left(\int_0^1 (x^{2k-2}-x^{2k-1})\,dx\right)\\\\ &=\int_0^1 \sum_{k=1}^n \left(x^{2k-2}-x^{2k-1}\right)\,dx\\\\ &=\int_0^1\frac{1-x^{2n}}{1+x}\,dx \end{align}$$

Using the Dominated Convergence Theorem, we have

$$\begin{align} \lim_{n\to \infty}\sum_{k=n+1}^{2n}\frac1k&=\int_0^1\lim_{n\to \infty}\left(\frac{1-x^{2n}}{1+x}\right)\,dx\\\\ &=\int_0^1 \frac{1}{1+x}\,dx\\\\ &=\log(2) \end{align}$$

and we are done!

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