# 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 ?

• You cannot separate an infinite sum. Commented May 31, 2018 at 14:42
• can you explain why ? Commented May 31, 2018 at 14:53
• Well, one of the most famous counterexample is exactly the question you are asking. Commented May 31, 2018 at 14:56
• When you take the limit you create an infinite number of infinitesimal terms, The sum is indeterminate. Commented May 31, 2018 at 14:57

$$\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} + \frac{n}{n^2 +n}=\lim_{n\to\infty} \frac{\sum_{k=1}^n k}{n^2+n}=\lim_{n\to\infty} \frac{\sum_{k=1}^{n+1} k-\sum_{k=1}^n k}{(n+1)^2+(n+1)-n^2-n}=\lim_{n\to\infty} \frac{n+1}{2n+2}=\frac12$$
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: $$\sum_{k=1}^n \frac{k}{n^2+n} = \underbrace{\frac{1}{1+1/n}}_{\to\, 1} \cdot\underbrace{\sum_{k=1}^n x_k \Delta x}_{\to\,\int_0^1x\;dx \,=\,\frac12} \to \frac12$$