Question Number 46840 by maxmathsup by imad last updated on 01/Nov/18
$${find}\:{lim}_{{n}\rightarrow+\infty} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{n}^{\mathrm{2}\:} +{k}^{\mathrm{2}} }{{n}^{\mathrm{3}} \:+{k}^{\mathrm{3}} } \\ $$
Commented by maxmathsup by imad last updated on 03/Nov/18
![let S_n =Σ_(k=1) ^n ((n^(2 ) +k^2 )/(n^3 +k^3 )) ⇒ S_n =Σ_(k=1) ^n ((n^2 (1+(k^2 /n^2 )))/(n^3 (1+(k^3 /n^3 )))) =Σ_(k=1) ^n ((1+((k/n))^2 )/(1+((k/n))^3 )) so S_n is a Rieman sum and lim_(n→+∞) S_n = ∫_0 ^1 ((1+x^2 )/(1+x^3 ))dx =∫_0 ^1 (dx/(1+x^3 )) +∫_0 ^1 (x^2 /(1+x^3 )) dx but ∫_0 ^1 (x^2 /(1+x^3 ))dx =[(1/3)ln(1+x^3 )]_0 ^1 =((ln(2))/3) let decompose F(x)=(1/(x^3 +1)) =(1/((x+1)(x^2 −x +1))) =(a/(x+1)) +((bx +c)/(x^2 −x+1)) a =lim_(x→−1) (x+1)F(x) =(1/3) lim_(x→+∞) xF(x) =0 =a+b ⇒b =−(1/3) ⇒F(x)=(1/(3(x+1))) +((−(1/3)x +c)/(x^2 −x+1)) F(0) =1 =(1/3) +c ⇒c =(2/3) ⇒F(x)=(1/(3(x+1))) −(1/(3 )) ((x−2)/(x^2 −x +1)) ⇒ ∫_0 ^1 F(x) =(1/3)[ln∣x+1∣]_0 ^1 −(1/6)∫_0 ^1 ((2x−1−3)/(x^2 −x +1))dx =((ln(2))/3) −(1/6)[ln∣x^2 −x +1∣]_0 ^1 +(1/2) ∫_0 ^1 (dx/(x^2 −x +1)) =((ln(2))/3) +(1/2) ∫_0 ^1 (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) ((ln(2))/3) +(1/2) ∫_(−(1/( (√3)))) ^(1/( (√3))) (4/3) (1/(t^2 +1)) ((√3)/2)dt =((ln(2))/3) +(2/( (√3))) arctan((1/( (√3)))) =((ln(2))/3) +(2/( (√3))) (π/6) =((ln(2))/3) +(π/(3(√3))) ⇒ lim_(n→+∞) S_n =((2ln(2))/3) +(π/(3(√3))) .](https://www.tinkutara.com/question/Q47008.png)
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}^{\mathrm{2}\:} +{k}^{\mathrm{2}} }{{n}^{\mathrm{3}} \:+{k}^{\mathrm{3}} }\:\:\Rightarrow\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)}{{n}^{\mathrm{3}} \left(\mathrm{1}+\frac{{k}^{\mathrm{3}} }{{n}^{\mathrm{3}} }\right)}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{3}} } \\ $$$${so}\:{S}_{{n}} \:{is}\:{a}\:{Rieman}\:{sum}\:{and}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }\:{dx}\:{but}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}\:+\mathrm{1}\right)}\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}\:+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}\:+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{3}}\:+{c}\:\Rightarrow{c}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{3}\:}\:\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\left[{ln}\mid{x}+\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{3}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}{dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{6}}\left[{ln}\mid{x}^{\mathrm{2}} −{x}\:+\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\frac{\pi}{\mathrm{6}}\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} =\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$