Limit of a sum by integral: $\lim_{n\rightarrow\infty} \sum\limits_{i=1}^{n-1}\frac{i}{n^2}$

Question

I have to find the following limit by using an integral, but I have no idea what to do.

$\lim_{n\rightarrow\infty}(\sum\limits_{i=1}^{n-1}\frac{i}{n^2})$

I know that $\sum\limits_{i=1}^n\int_a^bf_i(x)dx = \int_a^b \sum\limits_{i=1}^\infty f(x)dx$ if $S_n\rightarrow f$ uniformly, but I don't know if that can help me here.

Thanks

Some other posts about the same limit (although without explicit requirement to use integral): Compute $\lim_{n\to \infty} \left({1 \over n^2} + {2 \over n^2} + \cdots + {n - 1 \over n^2}\right)$ and Evaluating the limit $\lim_{n\to\infty} \left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right)$ — Martin Sleziak, May 28 '18 at 10:53

score 2 · Accepted Answer · answered Nov 15 '14 at 18:29

2

Recall that from the definition of Riemann integral over a bounded interval, we have $$\lim_{n \to \infty} \dfrac{b-a}n \sum_{i=0}^{n-1} f(x_i) = \int_a^b f(x)$$ where $x_i = a +i \cdot \dfrac{b-a}n$. Hence, $$\lim_{n\to\infty}\sum_{i=1}^{n-1} \dfrac{i}{n^2}=\lim_{n\to\infty}\dfrac1n \sum_{i=1}^{n-1} \dfrac{i}n = \int_0^1 xdx = \dfrac12$$

answered Nov 15 '14 at 18:29

Adhvaitha

5,441

Could you please explain why? – Fraser Elliott Nov 15 '14 at 18:30
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@FraserElliott The first step is to remember the Riemann Integral Definition, then it's just do the computations, what is your doubt? What didn't you understanding? – DiegoMath Nov 15 '14 at 18:36

Limit of a sum by integral: $\lim_{n\rightarrow\infty} \sum\limits_{i=1}^{n-1}\frac{i}{n^2}$

1 Answers1