Limit problem with summation: $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \dots + \frac{n}{n^2 +n}$

Question

$\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} + \dots + \frac{n}{n^2 +n}$ 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 ?

Answer 1

By Stolz-Cesaro

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

Answer 2

Two other approaches:

1.$$\sum_{k=1}^n \frac{k}{n^2+n} = \frac{\sum_{k=1}^n k}{n^2+n} = \frac{n(n+1)/2}{n^2+n} =\frac{1}{2}$$

2. Let $\Delta x = \frac{1}{n}$ and $x_k=k\Delta x$, then we rewrite as a right Riemann sum: $$\sum_{k=1}^n \frac{k}{n^2+n} = \underbrace{\frac{1}{1+1/n}}_{\to\, 1} \cdot\underbrace{\sum_{k=1}^n x_k \Delta x}_{\to\,\int_0^1x\;dx \,=\,\frac12} \to \frac12 $$