# Convergence of the sequence $x_n= \frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$

$$\left\{x_n\right\}$$ is a convergent sequence where $$x_n= \frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$$. What is $$\lim\limits_{n\to\infty}x_n?$$

Here are my two approaches:

Using Euler's constant($$\gamma$$):

\begin{align}\lim_{n\rightarrow\infty}x_n &=\lim_{n\rightarrow\infty}(\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n})\\ &=\lim_{n\rightarrow\infty}[(1+\frac{1}{2}+\dots+\frac{1}{2n})-(1+\frac{1}{2}+...+\frac{1}{n})]\\ &=\lim_{n\rightarrow\infty}[(\gamma_{2n}+\log 2n)-(\gamma_n+\log n)]\\ &=\lim_{n\rightarrow\infty}(\gamma_{2n}-\gamma_n+\log 2)\\ &=\log(2) \end{align} because $$\lim_{n\rightarrow\infty}\gamma_{2n}=\gamma$$ and $$\lim_{n\rightarrow\infty}\gamma_n=\gamma$$.

2nd method:

\begin{align}x_n&=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\\ &=\frac{1}{n}(\frac{n}{n+1}+\frac{n}{n+2}+...+\frac{n}{2n})\\ &=\frac{1}{n}\sum_{i=1}^{n}\frac{n}{n+i} \end{align}

Now, as $$n\rightarrow\infty$$, each $$\frac{n}{n+i}\rightarrow1$$, hence $$\sum_{i=1}^{n}\frac{n}{n+i}\rightarrow n$$. Hence $$\lim_{n\rightarrow\infty}x_n=1$$.

Thus, from the two methods, I end up having two different answers. I suspect that the second method is wrong, but I cannot identify where the mistake is. And I also want to know a method to solve the sum without using Euler's constant. So if anyone can point out the mistake in the 2nd method, and modify it, then it will be of great help to me.

• What is wrong with method 2: You take the limit "by parts". First each term in that sum (and this is already wrong as the number of terms in that sum already depends on $\;n\;$ itself!), then you take the whole sum's limits...and then you multiply this with $\;\frac1n\;$ ....You can't do that! Jun 8, 2019 at 10:04
• Okay, I understand now. Jun 8, 2019 at 10:09

Yes, the second method is wrong: you can not apply the rule "the limit of a sum is the sum of limits" when the sum has a non constant number of terms. Note that $$x_n=\frac{1}{n}\left(\frac{n}{n+1}+\frac{n}{n+2}+...+\frac{n}{2n}\right)= \frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}$$ is a Riemann sum related to the integral $$\int_{0}^1\frac{1}{1+x}\,dx=\ln(2).$$ Therefore again we obtain that $$\lim_{n\rightarrow\infty}x_n=\ln(2).$$

Justify the following:

$$\sum_{k=1}^n\frac1{n+k}=\frac1n\sum_{k=1}^n\frac1{1+\frac kn}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{1+x}=\log2$$

Observe that you can rewrite $$x_n$$ as,

$$x_n =\frac{1}{n+1}+\frac{1}{n+1}+\dots+\frac{1}{2n}=\left(1+\frac{1}{2}+\frac{1}{3}+\dots \frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n}\right)$$

$$=\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2n}\right)$$

$$=\left(1-\frac{1}{2}+\frac{1}{3}+\dots -\frac{1}{2n}\right)$$

$$\lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots -\frac{1}{2n}\right) = \ln 2$$

The sequence limit exists otherwise the Taylor expansion of $$\ln 2$$ wouldn't exist.

Using another approach with harmonic numbers

$$x_n=\sum_{i=1}^n \frac 1{n+i}=H_{2 n}-H_n$$ Using asymptotics and Taylor expansions $$x_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$ We could do the same for $$y_n=\sum_{i=1}^{kn} \frac 1{n+i}=H_{(k+1)n}-H_n$$ and get $$y_n=\log (k+1)-\frac{k}{2 (k+1) n}+\frac{k (k+2)}{12 (k+1)^2 n^2}+O\left(\frac{1}{n^4}\right)$$