Limit of an arithmetic sequence

Question

$$\lim_{n \to \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+...+\frac{n-1}{n^2}\right)$$

$$S_n=\frac{n}{2}\cdot\left(\frac{2}{n^2}+(n-1)\cdot\frac{1}{n^2}\right)=\frac{n}{2}\left(\frac{2}{n^2}+\frac{n-1}{n^2}\right)=\frac{n}{2}\cdot\frac{n+1}{n^2}=\frac{n^2+n}{2n^2}$$

$$\lim_{n \to \infty}\frac{n^2+n}{2n^2}=\lim_{n \to \infty}\frac{\frac{n^2}{n^2}+\frac{n}{n^2}}{\frac{2n^2}{n^2}}=\lim_{n \to \infty}\frac{1+\frac{1}{n}}{2}=\frac{1}{2}$$

Is it correct? is there a way to use the squeeze theorem?

Are you aware of the Riemann Sum technique for proving such limits? — user236182, Nov 05 '15 at 21:04
It is correct. There are other more complicated ways. For instance our sum is a Riemann sum. — André Nicolas, Nov 05 '15 at 21:05
The limit is equal to $\lim_{n\to \infty}\left(\frac{1}{n}\left(\sum_{i=1}^{n-1}\frac{i}{n}\right)\right)=\int_0^1 x, dx=\frac{1}{2}$. — user236182, Nov 05 '15 at 21:23
Another question about the same limit: Compute $\lim_{n\to \infty} \left({1 \over n^2} + {2 \over n^2} + \cdots + {n - 1 \over n^2}\right)$ — Martin Sleziak, May 28 '18 at 10:47

score 1 · Accepted Answer · answered Nov 05 '15 at 21:08

Notice, your answer is correct. Here is a simple approach using sum of A.P.

$$\lim_{n\to \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots +\frac{n-1}{n^2}\right)$$ $$=\lim_{n\to \infty}\frac{1}{n^2}\left(1+2+3+\ldots +(n-1)\right)$$ $$=\lim_{n\to \infty}\frac{1}{n^2}\left(\frac{n(n-1)}{2}\right)$$$$=\frac{1}{2}\lim_{n\to \infty}\left(\frac{n-1}{n}\right)$$ $$=\frac{1}{2}\lim_{n\to \infty}\left(1-\frac{1}{n}\right)=\frac{1}{2}\left(1-0\right)=\color{red}{\frac{1}{2}}$$

Limit of an arithmetic sequence

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