Calculating the limit of the sum. I tried calculating the sum:
$$
\lim_{n\rightarrow ∞}\:( \frac{1}{n+1} +  \frac{1}{n+2} + \frac{1}{n+3}+.......+ \frac{1}{n+n})
$$
using the Sandwich Theorem, however only got that the limit is between $0.5$ and $1$, and was unable to go further with it...? Are there other approaches here? 
 A: Hint : 
$$\rm L = \int_{0}^1 \frac{1}{1+x} \rm dx$$
A: write
$$\lim_{n\rightarrow ∞} \left( \frac{1}{1+\frac1n} +  \frac{1}{1+\frac2n} + \frac{1}{1+\frac3n}+.......+ \frac{1}{1+\frac nn} \right)\frac1n$$
and use riemann sum.
A: $\displaystyle \sum_{k=1}^n \frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$
The limit is $\displaystyle \int_0^1\frac{1}{1+x}dx$
A: By Harmonic series
$$\sum_{k=1}^n \frac{1}{n+k}=\sum_{k=1}^{2n} \frac{1}{n}-\sum_{k=1}^n \frac{1}{n}\sim \ln 2n-\ln n= \ln 2$$
A: Note that
$$
\sum_{m=1}^k\frac{1}{m}\stackrel{\text{def}}=H_{k}=\log k+\gamma+o(1)
$$
where $\gamma$ is the Euler Mascheroni Constant. Hence
$$
\sum_{m=n+1}^{2n}\frac{1}{m}=H_{2n}-H_{n}=\log(2n)+\gamma-\log(n)-\gamma+o(1)\stackrel{n\to\infty}\rightarrow \log(2)
$$
since
$$
\log(2n)-\log(n)=\log 2.
$$
A: Other way.
We can show this identity
$$ \sum^{n}_{k=1} \frac{1}{k+n} =\sum^{2n}_{k=1}\frac{(-1)^{k+1}}{k} .   $$
So by the $ln$ series we have
$$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}=ln(2)$$
A: 
I thought it might be instructive to present an approach that relies only on straightforward arithmetic along with knowledge of the Taylor series $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$.


Note that we can write
$$\begin{align}
\sum_{k=1}^n \frac{1}{k+n}&=\sum_{k=n+1}^{2n}\frac1k\\\\
&=\sum_{k=1}^{2n}\frac1k -\sum_{k=1}^n\frac1k\\\\
&=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac1{2k}\right)-\sum_{k=1}^n\frac1k\\\\
&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\
&=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}\tag1
\end{align}$$
Using the Taylor series for $\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k}$, we see that 
$$\log(2)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\tag2$$
Finally, letting $n\to \infty$ in $(1)$, we see from $(2)$ that 
$$\lim_{n\to \infty}\sum_{k=1}^n \frac1{k+n}=\log(2)$$
