# Solve $\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$ without using Riemann sums.

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?

• $\sum_{k=1}^n \frac{1}{n+k} = \sum_{\ell=n+1}^{2n} \frac{1}{\ell}$. Can you use the asymptotics of the harmonic series as a known result? – Clement C. Feb 24 '16 at 17:47
• – Martin Sleziak Aug 31 '18 at 15:32

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 $$\ln(2)=\ln(2n)-\ln(n)\ge\sum_{k=1}^n\frac1{n+k}\ge\ln(2n+1)-\ln(n+1)=\ln\left(2-\frac1{n+1}\right)$$

• I took the liberty of correcting a typo in the second line; I hope that you don't mind. - Mark – Mark Viola Feb 24 '16 at 21:00
• Danke. I had the "+k" inserted in many places, but forgot that one. – LutzL Feb 25 '16 at 0:02
• Nice! Simple and clever answer. Thank you :) – user281593 Feb 25 '16 at 16:38

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 $n \to \infty$.

• How do you get that $\lim_{n \to \infty} H_n - \log n - \gamma = 0$ without using Riemann sums? – Michael Lugo Feb 24 '16 at 18:37
• @MichaelLugo The Euler-Maclaurin summation formula provides the expansion for the Harmonic number. – Mark Viola Feb 25 '16 at 17:53

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 $x=1$.

• 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? – Clement C. Feb 24 '16 at 18:18
• Yes, you are right. – detnvvp Feb 24 '16 at 18:22

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!