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Question Number 42098 by abdo.msup.com last updated on 17/Aug/18

$$let{S}_{n,p}=\sum _{k=1}^n\frac{1}{\sqrt{k+p}}$$$$findaequivalentof{S}_{n,p}whenn\to +\mathrm{\infty }$$$$pintegr⩾1.$$

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

$$thesequence{\left(\frac{1}{\sqrt{k}}\right)}_{k⩾1}isdecreasing\Rightarrow {\int }_k^{k+1}\frac{dt}{\sqrt{t+p}}⩽\frac{1}{\sqrt{k+p}}⩽{\int }_{k-1}^k\frac{dt}{\sqrt{t+p}}\Rightarrow$$$$\sum _{k=1}^n{\int }_k^{k+1}\frac{dt}{\sqrt{t+p}}⩽\sum _{k=1}^n\frac{1}{\sqrt{k+p}}⩽\sum _{k=1}^n{\int }_{k-1}^k\frac{dt}{\sqrt{t+p}}\Rightarrow$$$${\int }_1^{n+1}\frac{dt}{\sqrt{t+p}}⩽S_{n,p}⩽{\int }_0^n\frac{dt}{\sqrt{t+p}}\Rightarrow {\left[2\sqrt{t+p}\right]}_1^{n+1}⩽S_{n,p}⩽{\left[2\sqrt{t+p}\right]}_0^n\Rightarrow$$$$2\sqrt{n+1+p}-2\sqrt{p+1}⩽S_{n,p}⩽2\sqrt{n+p}-2\sqrt{p}\Rightarrow$$$$\frac{2\sqrt{n+1+p}-2\sqrt{p+1}}{2\sqrt{n+p}}⩽\frac{S_{n,p}}{2\sqrt{n+p}}⩽\frac{2\sqrt{n+p}-2\sqrt{p}}{2\sqrt{n+p}}\Rightarrow$$$$\sqrt{1+\frac{1}{n+p}}-\sqrt{\frac{p+1}{n+p}}⩽\frac{S_{n,p}}{2\sqrt{n+p}}⩽1-\sqrt{\frac{p}{n+p}}but$$$${lim}_{n\to +\mathrm{\infty }}\sqrt{1+\frac{1}{n+p}}-\sqrt{\frac{p+1}{n+1}}=1and{lim}_{n\to +\mathrm{\infty }}1-\sqrt{\frac{p}{n+p}}=1\Rightarrow$$$$S_{n,p}\sim 2\sqrt{n+p}(n\to +\mathrm{\infty }andpfixed)$$$$$$

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