Limit of this sum 
Evaluate $\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{n+n})$.

I converted the limit of this sum to the sum of limits. Then in each term, I divided the numerator and denominator by $n$. Each limit came out to be zero. Hence I got the answer as $0$. Have I gone wrong somewhere or is my solution correct? I have mixed feelings. Please help.

Ok: This is inspired on Riemann Sums
Let $S_n=\sum_{k=0}^{n-1}{\frac{1}{n+k+1}}$ be your sum
We first split $[0;1]$ into $n$ intervals of the form $[\frac{k}{n};\frac{k+1}{n}]$, with $k\in\{0,1,2,...,n-1\}$ 
Note that if $\frac{k}{n}\le x\le \frac{k+1}{n}$ then 
$\frac{1}{\frac{k+1}{n}+1}\le\frac{1}{1+x}\le\frac{1}{1+\frac{k}{n}}$. Now we integrate:
$$\frac{1}{k+n+1}=\int_{\frac{k}n}^{\frac{k+1}{n}}\frac{dx}{\frac{k+1}{n}+1}\le\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+x}\le\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+\frac{k}{n}}=\frac{1}{k+n}$$ thus:
$$\sum_{k=0}^{n-1}{\frac{1}{k+n+1}}\le \sum_{k=0}^{n-1}\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+x}\le\sum_{k=0}^{n-1}{\frac{1}{k+n}}$$ 
Which means:
$$S_n\le\int_{0}^{1}{\frac{dx}{x+1}}\le S_n-\frac{1}{n}+\frac{1}{n+n}$$
Now take the limit of both side
