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Question Number 212627 by MrGaster last updated on 19/Oct/24

$$$$$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{\sqrt{1\cdot 2}}{n^2+1}+\frac{\sqrt{2\cdot 3}}{n^2+2}+\dots +\frac{\sqrt{n(n+1)}}{n^2+n})$$

Answered by mehdee7396 last updated on 19/Oct/24

$$\frac{1}{n^2+n}+\frac{2}{n^2+n}+\frac{3}{n^2+n}+...+\frac{n}{n^2+n}<S_n$$$$\Rightarrow \frac{n(n+1)}{2(n^2+n)}<S_n$$$$S_n<\frac{2}{n^2}+\frac{3}{n^2}+\frac{4}{n^2}+...+\frac{n+1}{n^2}$$$$\Rightarrow S_n<\frac{n(n+3)}{2n^2}$$$$\Rightarrow \frac{n(n+1)}{2(n^2+n)}<S_n<\frac{n(n+3)}{2n^2}$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}{S}_n=\frac{1}{2}✓$$$$$$

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