Question Number 71649 by TawaTawa last updated on 18/Oct/19
Commented by mathmax by abdo last updated on 18/Oct/19
![Σ_(n=1) ^([x]) (√n)=1+(√2)+(√3)+...+(√([x])) (2/3)(√((x+(1/2))^3 ))=(2/3)(√(x^3 (1+(1/(2x)))^3 ))=(2/3)x^(3/2) (√((1+(1/(2x)))^3 ))∼(2/3)x^(3/2) let find an ∼ for f(x)=Σ_(k=1) ^([x]) (√k) we have 1+(√2)+(√3)+...+(√([x]))=(√([x))]{(1/(√([x]))) +(√(2/([x])))+(√(3/([x])))+...+(√(([x]−1)/([x])))} [x]=N =(√([N])){Σ_(k=1) ^(N−1) (√(k/N))}=N(√N)×(1/N)Σ_(k=1) ^(N−1) (√(k/N)) but (1/N)Σ_(k=1) ^(N−1) (√(k/N))=∫_0 ^1 (√x)dx =(2/3)x^(3/3) ]_0 ^1 =(2/3) ⇒ 1+(√2)+(√3)+...+(√([x]))∼(2/3)[x](√([x])) =(2/3)[x]^(3/2) ∼(2/3)x^(3/2) lim_(x→+∞) ((2/3)(√((x+(1/2))^3 ))−Σ_(n=1) ^x (√n)) =0 and if you have the answer post it .](Q71697.png)
$$\sum_{{n}=\mathrm{1}} ^{\left[{x}\right]} \sqrt{{n}}=\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+...+\sqrt{\left[{x}\right]} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{3}} }=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \sqrt{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{3}} }\sim\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${let}\:{find}\:{an}\:\sim\:{for}\:{f}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\left[{x}\right]} \sqrt{{k}}\:\:{we}\:{have} \\ $$$$\left.\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+...+\sqrt{\left[{x}\right]}=\sqrt{\left[{x}\right.}\right]\left\{\frac{\mathrm{1}}{\sqrt{\left[{x}\right]}}\:+\sqrt{\frac{\mathrm{2}}{\left[{x}\right]}}+\sqrt{\frac{\mathrm{3}}{\left[{x}\right]}}+...+\sqrt{\frac{\left[{x}\right]−\mathrm{1}}{\left[{x}\right]}}\right\} \\ $$$$\left[{x}\right]={N}\:=\sqrt{\left[{N}\right]}\left\{\sum_{{k}=\mathrm{1}} ^{{N}−\mathrm{1}} \sqrt{\frac{{k}}{{N}}}\right\}={N}\sqrt{{N}}×\frac{\mathrm{1}}{{N}}\sum_{{k}=\mathrm{1}} ^{{N}−\mathrm{1}} \sqrt{\frac{{k}}{{N}}} \\ $$$$\left.{but}\:\frac{\mathrm{1}}{{N}}\sum_{{k}=\mathrm{1}} ^{{N}−\mathrm{1}} \sqrt{\frac{{k}}{{N}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}}{dx}\:=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+...+\sqrt{\left[{x}\right]}\sim\frac{\mathrm{2}}{\mathrm{3}}\left[{x}\right]\sqrt{\left[{x}\right]}\:=\frac{\mathrm{2}}{\mathrm{3}}\left[{x}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} \:\sim\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${lim}_{{x}\rightarrow+\infty} \:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }−\sum_{{n}=\mathrm{1}} ^{{x}} \sqrt{{n}}\right)\:=\mathrm{0} \\ $$$${and}\:{if}\:{you}\:{have}\:{the}\:{answer}\:{post}\:{it}\:. \\ $$
Commented by TawaTawa last updated on 19/Oct/19
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{will}\:\mathrm{use}\:\mathrm{your}\:\mathrm{answer}.\:\mathrm{God}\:\mathrm{bless} \\ $$$$\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Abdo msup. last updated on 19/Oct/19
$${you}\:{are}\:{most}\:{welcome}. \\ $$