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Question Number 42096 by abdo.msup.com last updated on 17/Aug/18

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

Commented by maxmathsup by imad last updated on 18/Aug/18

$$thesequence{\left(\frac{1}{\sqrt{k}}\right)}_{k⩾1}isdecreasingso$$$${\int }_k^{k+1}\frac{dt}{\sqrt{t}}⩽\frac{1}{\sqrt{k}}⩽{\int }_{k-1}^k\frac{dt}{\sqrt{t}}\Rightarrow \sum _{k=1}^n{\int }_k^{k+1}\frac{dt}{\sqrt{t}}⩽\sum _{k=1}^n\frac{1}{\sqrt{k}}⩽\sum _{k=1}^n{\int }_{k-1}^k\frac{dt}{\sqrt{t}}\Rightarrow$$$${\int }_1^{n+1}\frac{dt}{\sqrt{t}}⩽S_n⩽{\int }_0^n\frac{dt}{\sqrt{t}}\Rightarrow {\left[2\sqrt{t}\right]}_1^{n+1}⩽S_n⩽{\left[2\sqrt{t}\right]}_1^n\Rightarrow$$$$2\sqrt{n+1}-2⩽S_n⩽2\sqrt{n}\Rightarrow \frac{2\sqrt{n+1}-2}{2\sqrt{n}}⩽\frac{S_n}{2\sqrt{n}}⩽1\Rightarrow$$$$\sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}⩽\frac{S_n}{2\sqrt{n}}⩽1butwehave{lim}_{n\to +\mathrm{\infty }}\sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}=1\Rightarrow$$$$S_n\sim 2n(n\to +\mathrm{\infty })$$$$$$

Commented by maxmathsup by imad last updated on 18/Aug/18

$$S_n\sim 2\sqrt{n}(n\to +\mathrm{\infty })$$

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