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Question Number 57409 by Abdo msup. last updated on 03/Apr/19

$$let{S}_n=\sum _{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}$$$$calculste{lim}_{n\to +\mathrm{\infty }}{S}_n$$

Commented by maxmathsup by imad last updated on 10/Apr/19

$$wehave\frac{1}{\sqrt{n^2+k}}=\frac{1}{n\sqrt{1+\frac{k}{n^2}}}=\frac{{(1+\frac{k}{n^2})}^{-\frac{1}{2}}}{n}$$$$butwehave{(1+u)}^{\alpha }=1+\alpha u+\frac{\alpha (\alpha -1)}{2}{u}^2+0\left(u^3\right)\Rightarrow$$$${(1+\alpha )}^{-\frac{1}{2}}=1-\frac{\alpha }{2}+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2}{\alpha }^2+o\left({\alpha }^3\right)$$$$=1-\frac{\alpha }{2}+\frac{3}{8}{\alpha }^2+0\left({\alpha }^3\right)\Rightarrow 1-\frac{\alpha }{2}⩽{(1+\alpha )}^{-\frac{1}{2}}⩽1-\frac{\alpha }{2}+\frac{3}{8}{\alpha }^2\Rightarrow$$$$1-\frac{k}{2n^2}⩽{(1+\frac{k}{n^2})}^{-\frac{1}{2}}⩽1-\frac{k}{2n^2}+\frac{3}{8}\frac{k^2}{n^4}\Rightarrow \frac{1}{n}-\frac{k}{2n^3}⩽\frac{1}{n}{(1+\frac{k}{n^2})}^{-\frac{1}{2}}⩽$$$$\frac{1}{n}-\frac{k}{2n^3}+\frac{3}{8}\frac{k^2}{n^5}\Rightarrow \sum _{k=1}^{2n+1}(\frac{1}{n}-\frac{k}{2n^3})⩽S_n⩽\sum _{k=1}^{2n+1}(\frac{1}{n}-\frac{k}{2n^3}+\frac{3}{8}\frac{k^2}{n^5})\Rightarrow$$$$2-\frac{1}{2n^3}\sum _{k=1}^{2n+1}k⩽S_n⩽2-\frac{1}{2n^3}\sum _{k=1}^{2n+1}k+\frac{3}{8n^5}\sum _{k=1}^{2n+1}{k}^2$$$$2-\frac{1}{2n^3}\frac{(2n+1)(2n+2)}{2}⩽S_n⩽2-\frac{1}{2n^3}\frac{(2n+1)(2n+2)}{2}+\frac{3}{8n^5}\frac{(2n+1)(2n+2)(4n+2+1)}{6}$$$$\Rightarrow 2-\frac{(n+1)(2n+1)}{2n^3}⩽S_n⩽2-\frac{(n+1)(2n+1)}{2n^3}+\frac{1}{8n^5}(n+1)(2n+1)(4n+3)$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=2.$$$$$$

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