find-the-limit-lim-n-sin-1-n-2-sin-2-n-2-sin-n-n-2-

Question Number 124355 by Eric002 last updated on 02/Dec/20

f i n d t h e l i m i t

\underset{n \to \infty}{l i m} (s i n (\frac{1}{n^{2}}) + s i n (\frac{2}{n^{2}}) + \dots \dots + s i n (\frac{n}{n^{2}}))

Answered by Dwaipayan Shikari last updated on 02/Dec/20

\lim_{n \to \infty} s i n (\frac{1}{n^{2}}) + s i n (\frac{2}{n^{2}}) + \dots

= \frac{1 + 2 + 3 + 4 + 5 + 6 + . . + n}{n^{2}} = \frac{n^{2} + n}{2 n^{2}} = \frac{1}{2} (A s s i n (\frac{1}{n^{2}}) \to (\frac{1}{n^{2}}))

Answered by mathmax by abdo last updated on 02/Dec/20

we have x - \frac{x^{3}}{6} ⩽ sinx ⩽ x \Rightarrow \sum_{k = 1}^{n} \frac{k}{n^{2}} - \frac{1}{6} \sum_{k = 1}^{n} \frac{k^{3}}{n^{6}} ⩽ \sum_{k = 1}^{n} \sin (\frac{k}{n^{2}}) ⩽ \sum_{k = 1}^{n} \frac{k}{n^{2}} \Rightarrow

\frac{n (n + 1)}{{2 n}^{2}} - \frac{1}{{6 n}^{3}} {(\frac{n (n + 1)}{2})}^{2} ⩽ S_{n} ⩽ \frac{n (n + 1)}{{2 n}^{2}} we passe to [limit (n \to \infty)

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