Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$
the only thing I could think of is that it can be written like this :
$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$
is my answer right ? and are there any other method ?(I'm sure there are)
$$\int_{k}^{k+1} \frac{1}{x}dx \leq \dfrac{1}{k} \leq \int_{k-1}^{k} \frac{1}{x}dx.$$ $$ \ln\frac{2n+1}{n} \leq \sum_{k=n}^{2n}\frac{1}{k} \leq \ln\frac{2n}{n-1}. $$
We are going to use the Euler's constant $$\lim_{n\to\infty}\left(\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}-\ln (2n)\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\right)\right)=\lim_{n\to\infty}(\gamma_{2n}-\gamma_{n})=0$$
Hence the limit is $\ln 2$.
The function $f(x)=\frac{1}{1+x}$ is continuous on $[0,1]$, hence the Riemann sum converges to the integral:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac kn}= \int_0^1\frac{\mathrm dx}{1+x}=\ln 2$$
$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n+k}=\ln 2$$
$$\lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$$
I thought it might be instructive to present an approach that uses straightforward arithmetic and application of Leibniz's Test for alternating series. To that end, we now proceed.
It is not difficult to show that
$$\sum_{k=n+1}^{2n} \frac{1}{k}=\sum_{k=1}^{2n }\frac{(-1)^{k-1}}{k}\tag1$$
To show the validity of $(1)$, we simply write
$$\begin{align} \sum_{k=n+1}^{2n}\frac1k&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n \frac1k\\\\ &=\sum_{k=1}^n\left(\frac1{2k}+\frac1{2k+1}\right)-2\sum_{k=1}^n\frac1{2k}\\\\ &=\sum_{k=1}^n\left(\frac1{2k+1}-\frac1{2k}\right)\\\\ &=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}\tag2 \end{align}$$
Notice that the sum on the right-hand side is the alternating harmonic series, which converges as guaranteed by Leibniz's test. And we conclude that the series converges!
BONUS: EVALUATING THE LIMIT
In fact, using the Taylor series of $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$, and evaluating it at $x=1$, we find that the series of interest is equal to $\log(2)$. Hence, we find that
$$\lim_{n\to\infty}\sum_{k=n+1}^{2n} \frac{1}{k}=\log(2)$$
And we are done.
Set $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}.$$
We can use the following uniform bound $$a_n = \sum_{k=1}^n \frac{1}{n+k} \leq\sum_{k=1}^n\frac{1}{n+1} = \frac{n}{n+1} < 1$$ Since each $a_n$ is positive, we have that $0\leq a_n < 1$ for all $n$. For monotonicity we use the calculation $$\begin{align*}a_{n}-a_{n+1} &= \sum_{k=1}^n\frac{1}{n+k} - \sum_{k=1}^{n+1}\frac{1}{(n+1) + k}\\ &= \frac{1}{n+1} - \frac{1}{2n+1} - \frac{1}{2n+2}\\ &= \frac{(2n+1)(2n+2)- (n+1)(2n+2) - (n+1)(2n+1)}{(n+1)(2n+1)(2n+2)}\\ &= -\frac{(n+1)}{(n+1)(2n+1)(2n+2)}\\ &= -\frac{1}{(2n+1)(2n+2)} < 0\\ \end{align*}$$ We thus have that $\{a_n\}$ is bounded and the statement above shows that $a_n < a_{n+1}$ for any $n$ so that $\{a_n\}$ is also monotonic. Hence the sequence converges.
Set $a_n = 1/(n+1) + 1/(n+2)+···+ 1/(2n) =\sum_{k=1}^n \dfrac1{n+k} $.
$\begin{array}\\ a_{n+1}-a_n &=\sum_{k=1}^{n+1} \dfrac1{n+1+k}-\sum_{k=1}^n \dfrac1{n+k}\\ &=\sum_{k=2}^{n+2} \dfrac1{n+k}-\sum_{k=1}^n \dfrac1{n+k}\\ &=\dfrac1{n+n+2}+\dfrac1{n+n+1}-\dfrac1{n+1}\\ &=\dfrac1{2n+2}+\dfrac1{2n+1}-\dfrac1{n+1}\\ &=-\dfrac1{2 (n + 1) (2 n + 1)}\\ &< 0\\ \text{and}\\ a_{n+1}-a_n &=-\dfrac1{2 (n + 1) (2 n + 1)}\\ &>-\dfrac1{2n (n + 1)}\\ \text{so} &\text{if } m > 0\\ a_{m+n}-a_n &< 0\\ \text{and}\\ a_{m+n}-a_n &=\sum_{k=0}^{m-1}(a_{n+k+1}-a_{n+k})\\ &>-\sum_{k=0}^{m-1}(\dfrac1{2(n+k+1) (n + k+2)})\\ &=-\dfrac12\sum_{k=0}^{m-1}(\dfrac1{(n+k+1)}-\dfrac1{(n + k+2)})\\ &=-\dfrac12(\dfrac1{(n+1)}-\dfrac1{(n +m+1)})\\ &>-\dfrac1{2(n+1)}\\ \end{array} $
so, for $m > 0$, $|a_{m+n}-a_n| \lt \dfrac1{2(n+1)} \to 0 $ as $n \to \infty$ so, by Cauchy, $a_n$ converges.
