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Question Number 71649 by TawaTawa last updated on 18/Oct/19

Commented by mathmax by abdo last updated on 18/Oct/19

$$\sum _{n=1}^{\left[x\right]}\sqrt{n}=1+\sqrt{2}+\sqrt{3}+...+\sqrt{\left[x\right]}$$$$\frac{2}{3}\sqrt{{(x+\frac{1}{2})}^3}=\frac{2}{3}\sqrt{x^3{(1+\frac{1}{2x})}^3}=\frac{2}{3}{x}^{\frac{3}{2}}\sqrt{{(1+\frac{1}{2x})}^3}\sim \frac{2}{3}{x}^{\frac{3}{2}}$$$$letfindan\sim forf\left(x\right)=\sum _{k=1}^{\left[x\right]}\sqrt{k}wehave$$$$1+\sqrt{2}+\sqrt{3}+...+\sqrt{\left[x\right]}=\sqrt{[x}]\{\frac{1}{\sqrt{\left[x\right]}}+\sqrt{\frac{2}{\left[x\right]}}+\sqrt{\frac{3}{\left[x\right]}}+...+\sqrt{\frac{\left[x\right]-1}{\left[x\right]}}\}$$$$\left[x\right]=N=\sqrt{\left[N\right]}\left\{\sum _{k=1}^{N-1}\sqrt{\frac{k}{N}}\right\}=N\sqrt{N}\times \frac{1}{N}\sum _{k=1}^{N-1}\sqrt{\frac{k}{N}}$$$${but\frac{1}{N}\sum _{k=1}^{N-1}\sqrt{\frac{k}{N}}={\int }_0^1\sqrt{x}dx=\frac{2}{3}{x}^{\frac{3}{3}}]}_0^1=\frac{2}{3}\Rightarrow$$$$1+\sqrt{2}+\sqrt{3}+...+\sqrt{\left[x\right]}\sim \frac{2}{3}\left[x\right]\sqrt{\left[x\right]}=\frac{2}{3}{\left[x\right]}^{\frac{3}{2}}\sim \frac{2}{3}{x}^{\frac{3}{2}}$$$${lim}_{x\to +\mathrm{\infty }}(\frac{2}{3}\sqrt{{(x+\frac{1}{2})}^3}-\sum _{n=1}^x\sqrt{n})=0$$$$andifyouhavetheanswerpostit.$$

Commented by TawaTawa last updated on 19/Oct/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{the}\mathrm{answer}\mathrm{sir}.\mathrm{I}\mathrm{will}\mathrm{use}\mathrm{your}\mathrm{answer}.\mathrm{God}\mathrm{bless}$$$$\mathrm{you}\mathrm{sir}.$$

Commented by Abdo msup. last updated on 19/Oct/19

$$youaremostwelcome.$$

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