Limit problem with summation: $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \dots + \frac{n}{n^2 +n}$ $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} + \dots + \frac{n}{n^2 +n}$
question is when we take limit we can seperate things right ? So we can write $\lim_{n\to\infty} \frac{1}{n^2 + n}$ + $\lim_{n\to\infty} \frac{2}{n^2 + n}$ + .... $\lim_{n\to\infty} \frac{n}{n^2 + n}$
if we take limits one by one we get zeroes. 
we get sum = 0 
but if we do sum first than take limit
$\lim_{n\to\infty} \frac{\frac{n.(n+1)}{2}}{n^2 + n}$ with simplification we get 1/2 
so did my first question wrong ? can't we take limits first than do the sum ?

By Stolz-Cesaro
$$\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} +  \frac{n}{n^2 +n}=\lim_{n\to\infty} \frac{\sum_{k=1}^n k}{n^2+n}=\lim_{n\to\infty} \frac{\sum_{k=1}^{n+1} k-\sum_{k=1}^n k}{(n+1)^2+(n+1)-n^2-n}=\lim_{n\to\infty} \frac{n+1}{2n+2}=\frac12$$
