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Question Number 205772 by mr W last updated on 30/Mar/24

Answered by MM42 last updated on 30/Mar/24

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

Answered by A5T last updated on 30/Mar/24

$$T_0=\sqrt{2}-0-1$$$$T_{n-1}=\sqrt{n^2+n}-\sqrt{n^2-n}-1$$$$T_n=\sqrt{n^2+3n+2}-\sqrt{n^2+n}-1$$$$S_n=-(n+1)+\sqrt{n^2+3n+2}$$$$n+1+n+2>2\sqrt{(n+1)(n+2)}\Rightarrow \sqrt{n^2+3n+2}<\frac{2n+3}{2}$$$$\Rightarrow S_n=\sqrt{n^2+3n+2}-n-1<n+\frac{3}{2}-n-1=\frac{1}{2}$$$$\text{Double subscripts: use braces to clarify}$$

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