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Question Number 61041 by mathsolverby Abdo last updated on 28/May/19
calculate ∫_1 ^(+∞)  (([2x]−[x])/x^4 ) dx
$${calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[\mathrm{2}{x}\right]−\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\: \\ $$
Commented by maxmathsup by imad last updated on 29/May/19
let A =∫_1 ^(+∞)  (([2x]−[x])/x^4 ) dx ⇒ A =∫_1 ^(+∞)  (([2x])/x^4 ) dx −∫_1 ^(+∞)  (([x])/x^4 ) dx  ∫_1 ^(+∞)  (([2x])/x^4 ) dx =_(2x=t)   2^4   ∫_2 ^(+∞)  (([t])/t^4 )  (dt/2) =8 ∫_2 ^(+∞)  (([t])/t^4 ) dt  =8 ( ∫_1 ^(+∞)  (([t])/t^4 ) dt −∫_1 ^2   (([t])/t^4 ) dt) =8 ∫_1 ^(+∞)  (([t])/t^4 ) dt −8 ∫_1 ^2   (dt/t^4 )  ∫_1 ^2   (dt/t^4 ) =[(1/(−4+1)) t^(−4+1) ]_1 ^2  =[−(1/3) t^(−3) ]_1 ^2  =−(1/3)(2^(−3) −1)=(1/3)(1−(1/8))=(1/3) (7/8) ⇒  ∫_1 ^(+∞)  (([2x])/x^4 ) dx =8 ∫_1 ^(+∞)  (([t])/t^4 )dt−(7/3) ⇒A =7 ∫_1 ^(+∞)  (([x])/x^4 )dx −(7/3)  ∫_1 ^(+∞)    (([x])/x^4 ) dx =lim_(n→+∞)    Σ_(k=1) ^(n−1)  ∫_k ^(k+1)  (([x])/x^4 ) dx  =Σ_(n=1) ^∞   ∫_n ^(n+1)  (([x])/x^4 ) dx =Σ_(n=1) ^∞   n ∫_n ^(n+1)  (dx/x^4 ) =Σ_(n=1) ^∞ n[(1/(−4+1)) x^(−4+1) ]_n ^(n+1)   =Σ_(n=1) ^∞ n[−(1/3) (1/x^3 )]_n ^(n+1)  =−(1/3) Σ_(n=1) ^∞  n((1/((n+1)^3 )) −(1/n^3 ))  =(1/3) Σ_(n=1) ^∞  (1/n^2 ) −(1/3) Σ_(n=1) ^∞  (n/((n+1)^3 ))  =(1/3)ξ(2)−(1/3)Σ_(n=1) ^∞  ((n+1−1)/((n+1)^3 )) =(1/3)ξ(2)−(1/3) Σ_(n=1) ^∞  (1/((n+1)^2 )) +(1/3)Σ_(n=1) ^∞  (1/((n+1)^3 ))  Σ_(n=1) ^∞   (1/((n+1)^2 ))=Σ_(n=2) ^∞  (1/n^2 ) =ξ(2)−1  Σ_(n=1) ^∞  (1/((n+1)^3 )) =Σ_(n=2) ^∞  (1/n^3 ) =ξ(3)−1 ⇒  ∫_1 ^(+∞)  (([x])/x^4 ) dx = (1/3)ξ(2) −(1/3)(ξ(2)−1) +(1/3) (ξ(3)−1)  =(1/3) +(1/3)ξ(3)−(1/3) =(1/3)ξ(3) ⇒ A =(7/3)ξ(3)−(7/3) .
$${let}\:{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[\mathrm{2}{x}\right]−\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:\Rightarrow\:{A}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[\mathrm{2}{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:−\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[\mathrm{2}{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:=_{\mathrm{2}{x}={t}} \:\:\mathrm{2}^{\mathrm{4}} \:\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\left[{t}\right]}{{t}^{\mathrm{4}} }\:\:\frac{{dt}}{\mathrm{2}}\:=\mathrm{8}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\left[{t}\right]}{{t}^{\mathrm{4}} }\:{dt} \\ $$$$=\mathrm{8}\:\left(\:\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{t}\right]}{{t}^{\mathrm{4}} }\:{dt}\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\left[{t}\right]}{{t}^{\mathrm{4}} }\:{dt}\right)\:=\mathrm{8}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{t}\right]}{{t}^{\mathrm{4}} }\:{dt}\:−\mathrm{8}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} } \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} }\:=\left[\frac{\mathrm{1}}{−\mathrm{4}+\mathrm{1}}\:{t}^{−\mathrm{4}+\mathrm{1}} \right]_{\mathrm{1}} ^{\mathrm{2}} \:=\left[−\frac{\mathrm{1}}{\mathrm{3}}\:{t}^{−\mathrm{3}} \right]_{\mathrm{1}} ^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}^{−\mathrm{3}} −\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{7}}{\mathrm{8}}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[\mathrm{2}{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:=\mathrm{8}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{t}\right]}{{t}^{\mathrm{4}} }{dt}−\frac{\mathrm{7}}{\mathrm{3}}\:\Rightarrow{A}\:=\mathrm{7}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }{dx}\:−\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:={lim}_{{n}\rightarrow+\infty} \:\:\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{n}\:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{4}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\left[\frac{\mathrm{1}}{−\mathrm{4}+\mathrm{1}}\:{x}^{−\mathrm{4}+\mathrm{1}} \right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\left[−\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right]_{{n}} ^{{n}+\mathrm{1}} \:=−\frac{\mathrm{1}}{\mathrm{3}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{3}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\xi\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{3}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}+\mathrm{1}−\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{3}}\xi\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{3}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\xi\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi\left(\mathrm{3}\right)−\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\xi\left(\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\xi\left(\mathrm{2}\right)−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{3}}\:\left(\xi\left(\mathrm{3}\right)−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}\xi\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{3}}\xi\left(\mathrm{3}\right)\:\Rightarrow\:{A}\:=\frac{\mathrm{7}}{\mathrm{3}}\xi\left(\mathrm{3}\right)−\frac{\mathrm{7}}{\mathrm{3}}\:. \\ $$
Answered by perlman last updated on 28/May/19
∫_1 ^(+∞) (([2x]−[x])/x^4 )dx=Σ_(k=1) ^(+∞) ∫_((k+1)/2) ^((k+2)/2)    (([2x])/x^(4 ) )dx−Σ_(k=1) ^(+∞) ∫_k ^(k+1) (([x])/x^4 )dx  [2x]=k+1  ∀x∈[((k+1)/2),((k+2)/2)[  [x]=k  ∀x∈[k,k+1[  ∫_1 ^(+∞) (([2x]−[x])/x^4 )=lim_(n→∞) Σ_(k=1) ^n [∫_((k+1)/2) ^((k+2)/2) ((k+1)/x^4 )dx−∫_k ^(k+1) (k/x^(4 ) )dx]  =lim_(n→∞) Σ_(k=1) ^(k=n) [((k+1)/(−3(((k+2)/2))^3 ))+((k+1)/(3(((k+1)/2))^3 ))+(k/(3(k+1)^3 ))−(k/(3k^3 ))]  =lim_(n→∞) Σ_(k=1) ^(k=n) [((−8(k+1))/(   3(k+2)^3 ))+(8/(3(k+1)^2 ))+(k/(3(k+1)^3  ))−(1/(3k^2 ))]  we use  (x/((x+1)^3 ))=(1/((x+1)^2 ))−(1/((x+1)^3 )) andlim_(x→∞) Σ_(k=1) ^x (1/k^z )=ζ(z)  we find   limΣ_(k=1) ^n [((−8)/(3(k+2)^2 ))+(8/(3(k+2)^3 ))+(8/(3(k+1)^2 ))+(1/(3(k+1)^2 ))−(1/(3(k+1)^3 ))−(1/(3k^2 ))]  =lim [(8/(12))−(8/(3(n+2)^2 ))+(8/3)Σ_(k=1) ^n (1/((k+2)^3 ))+(1/(3(n+1)^2 ))−(1/3)−(1/3)Σ_(k=1) ^n (1/((k+1)^3 ))]  =(8/(12))−(1/3)+(8/3)Σ_(k=1) ^∞ (1/((k+2)^3 ))−(1/3)Σ_(k=1) ^∞ (1/((k+1)^3 ))=(1/3)+(8/3)(ζ(3)−1−(1/8))−(1/3)(ζ(3)−1)  =(1/3)+(7/3)ζ(3)−3+(1/3)=(7/3)ζ(3)−(7/3)=(7/3)(ζ(3)−1)
$$\int_{\mathrm{1}} ^{+\infty} \frac{\left[\mathrm{2}{x}\right]−\left[{x}\right]}{{x}^{\mathrm{4}} }{dx}=\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\int_{\frac{{k}+\mathrm{1}}{\mathrm{2}}} ^{\frac{{k}+\mathrm{2}}{\mathrm{2}}} \:\:\:\frac{\left[\mathrm{2}{x}\right]}{{x}^{\mathrm{4}\:} }{dx}−\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\underset{{k}} {\overset{{k}+\mathrm{1}} {\int}}\frac{\left[{x}\right]}{{x}^{\mathrm{4}} }{dx} \\ $$$$\left[\mathrm{2}{x}\right]={k}+\mathrm{1}\:\:\forall{x}\in\left[\frac{{k}+\mathrm{1}}{\mathrm{2}},\frac{{k}+\mathrm{2}}{\mathrm{2}}\left[\right.\right. \\ $$$$\left[{x}\right]={k}\:\:\forall{x}\in\left[{k},{k}+\mathrm{1}\left[\right.\right. \\ $$$$\underset{\mathrm{1}} {\overset{+\infty} {\int}}\frac{\left[\mathrm{2}{x}\right]−\left[{x}\right]}{{x}^{\mathrm{4}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\underset{\frac{{k}+\mathrm{1}}{\mathrm{2}}} {\overset{\frac{{k}+\mathrm{2}}{\mathrm{2}}} {\int}}\frac{{k}+\mathrm{1}}{{x}^{\mathrm{4}} }{dx}−\underset{{k}} {\overset{{k}+\mathrm{1}} {\int}}\frac{{k}}{{x}^{\mathrm{4}\:} }{dx}\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\sum}}\left[\frac{{k}+\mathrm{1}}{−\mathrm{3}\left(\frac{{k}+\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{3}} }+\frac{{k}+\mathrm{1}}{\mathrm{3}\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }+\frac{{k}}{\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{{k}}{\mathrm{3}{k}^{\mathrm{3}} }\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{k}={n}} {\sum}}\left[\frac{−\mathrm{8}\left({k}+\mathrm{1}\right)}{\:\:\:\mathrm{3}\left({k}+\mathrm{2}\right)^{\mathrm{3}} }+\frac{\mathrm{8}}{\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{k}}{\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{3}} \:}−\frac{\mathrm{1}}{\mathrm{3}{k}^{\mathrm{2}} }\right] \\ $$$${we}\:{use}\:\:\frac{{x}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:{and}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{k}^{{z}} }=\zeta\left({z}\right) \\ $$$${we}\:{find}\: \\ $$$${lim}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\frac{−\mathrm{8}}{\mathrm{3}\left({k}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}\left({k}+\mathrm{2}\right)^{\mathrm{3}} }+\frac{\mathrm{8}}{\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}{k}^{\mathrm{2}} }\right] \\ $$$$={lim}\:\left[\frac{\mathrm{8}}{\mathrm{12}}−\frac{\mathrm{8}}{\mathrm{3}\left({n}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{2}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }\right] \\ $$$$=\frac{\mathrm{8}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{7}}{\mathrm{3}}\zeta\left(\mathrm{3}\right)−\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{3}}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{3}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}\right) \\ $$$$ \\ $$
Commented by Tawa1 last updated on 28/May/19
Commented by Tawa1 last updated on 28/May/19
Please sir,  sum the series.
$$\mathrm{Please}\:\mathrm{sir},\:\:\mathrm{sum}\:\mathrm{the}\:\mathrm{series}.\:\: \\ $$
Commented by perlman last updated on 29/May/19
=Σ_(k=1) ^(+∞) (1/(k(k+1)(k+2)))  (1/(x(x+1)(x+2)))=(1/(2x))−(1/(x+1))+(1/(2(x+2)))  Σ_(k=1) ^(+∞) (1/(k(k+1)(k+2)))=lim_(x→+∞) Σ_(k=1) ^x [(1/(2k))−(1/(k+1))+(1/(2(k+2)))]  =lim_(x→+∞) Σ_(k=1) ^x (1/(2k))−Σ_(k=1) ^x (1/(k+1))+Σ(1/(2(k+2)))  put r=k+1 in the seconde sum  and v=k+2 in the 3rd  =lim_(x→+∞) Σ_(k=1) ^x (1/(2k))−Σ_(r=2) ^(x+1) (1/r)+Σ_(v=3) ^(x+2) (1/(2v))=lim_(x→+∞) ((1/2)+(1/4)+Σ_(k=3) ^x (1/(2k))−Σ_2 ^(x+1) (1/k)+Σ_3 ^x (1/(2k))+(1/(2x+2))+(1/(2(x+2))))  =lim_(x→+∞) ((3/4)+2Σ_(k=3) ^x (1/(2k))−(1/2)−Σ_(k=3) ^x (1/k)−(1/(x+1))+(1/(2x+2))+(1/(2x+4)))  =lim_(x→+∞) ((1/4)+Σ_3 ^x (1/k)−Σ_3 ^x (1/k)−(1/(2x+2))+(1/(2x+4)))  =lim_(x→+∞) ((1/4)−(1/(2x+2))+(1/(2x+4)))=(1/4)
$$=\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{2}{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{2}\right)}\right] \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}−\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}+\Sigma\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{2}\right)} \\ $$$${put}\:{r}={k}+\mathrm{1}\:{in}\:{the}\:{seconde}\:{sum} \\ $$$${and}\:{v}={k}+\mathrm{2}\:{in}\:{the}\:\mathrm{3}{rd} \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}−\underset{{r}=\mathrm{2}} {\overset{{x}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{r}}+\underset{{v}=\mathrm{3}} {\overset{{x}+\mathrm{2}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{v}}=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\underset{{k}=\mathrm{3}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}−\underset{\mathrm{2}} {\overset{{x}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}+\underset{\mathrm{3}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}\right) \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{2}\underset{{k}=\mathrm{3}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}−\frac{\mathrm{1}}{\mathrm{2}}−\underset{{k}=\mathrm{3}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{4}}\right) \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{4}}+\underset{\mathrm{3}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{k}}−\underset{\mathrm{3}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{4}}\right) \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 29/May/19
look sir  Σ_(k=1) ^n  (1/(k(k+1)(k+2))) =(1/(1.2.3)) +(1/(2.3.4)) +(1/(3.4.5)) +...so this isn t  the sum given at the question...!
$${look}\:{sir}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}\:+…{so}\:{this}\:{isn}\:{t} \\ $$$${the}\:{sum}\:{given}\:{at}\:{the}\:{question}…! \\ $$
Commented by perlman last updated on 29/May/19
2nd  un=(1/(n(n+1)(n+2)))  serie of therm un convergent  let f(x)=Σ_(n=1  ) ^(+∞) (x^(n+2) /(n(n+1)(n+2)))  f^((1)) (x)=Σ_(n=1) ^(+∞) (x^(n+1) /(n(n+1)))  f^((2)) (x)=Σ_(n=1) ^(+∞) (x^n /n)=−ln(1−x)“devlop of ln(1−z) in power series”  f^((1)) (x)=∫f^((1)) (x)dx=∫−ln(1−x)dx=−∫ln(1−x)dx=(1−x)ln(1−x)+x“caus f^((1)) (0)=0”  f(x)=∫f^((1)) dx=∫(1−x)ln(1−x)+x dx=((−(1−x)^2 ln(1−x))/2)+(((1−x)^2 )/4)+(x^2 /2)+c  f(0)=0===≥(1/4)+c=0    so c=−(1/4)  f(x)=((−(1−x)^2 ln(1−x))/2)+(((1−x)^2 )/4)+(x^2 /2)−(1/4)  for x=1 we get f(1)=Σ_(k=1) ^(+∞) (1/(k(k+1)(k+2)))=(1/2)−(1/4)=(1/4)
$$\mathrm{2}{nd} \\ $$$${un}=\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$${serie}\:{of}\:{therm}\:{un}\:{convergent} \\ $$$${let}\:{f}\left({x}\right)=\underset{{n}=\mathrm{1}\:\:} {\overset{+\infty} {\sum}}\frac{{x}^{{n}+\mathrm{2}} }{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{1}\right)} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{{x}^{{n}} }{{n}}=−{ln}\left(\mathrm{1}−{x}\right)“{devlop}\:{of}\:{ln}\left(\mathrm{1}−{z}\right)\:{in}\:{power}\:{series}'' \\ $$$${f}^{\left(\mathrm{1}\right)} \left({x}\right)=\int{f}^{\left(\mathrm{1}\right)} \left({x}\right){dx}=\int−{ln}\left(\mathrm{1}−{x}\right){dx}=−\int{ln}\left(\mathrm{1}−{x}\right){dx}=\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)+{x}“{caus}\:{f}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)=\mathrm{0}'' \\ $$$${f}\left({x}\right)=\int{f}^{\left(\mathrm{1}\right)} {dx}=\int\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)+{x}\:{dx}=\frac{−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}+\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}===\geqslant\frac{\mathrm{1}}{\mathrm{4}}+{c}=\mathrm{0}\:\:\:\:{so}\:{c}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${f}\left({x}\right)=\frac{−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}+\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${for}\:{x}=\mathrm{1}\:{we}\:{get}\:{f}\left(\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by perlman last updated on 29/May/19
y′re welcom
$${y}'{re}\:{welcom} \\ $$
Commented by Tawa1 last updated on 29/May/19
Wow. God bless you sir.  Sir,   Does the series has sum of   nth  term
$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{Sir},\:\:\:\mathrm{Does}\:\mathrm{the}\:\mathrm{series}\:\mathrm{has}\:\mathrm{sum}\:\mathrm{of}\:\:\:\boldsymbol{\mathrm{nth}}\:\:\mathrm{term} \\ $$
Commented by Tawa1 last updated on 29/May/19
I want to know if the sum to nth term correspond to the given series.  that is,   using the sum of nth term formular to verify.  S_1  ,  S_2  ,  S_3   ...  if it correspond
$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{if}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{correspond}\:\mathrm{to}\:\mathrm{the}\:\mathrm{given}\:\mathrm{series}. \\ $$$$\mathrm{that}\:\mathrm{is},\:\:\:\mathrm{using}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{formular}\:\mathrm{to}\:\mathrm{verify}.\:\:\mathrm{S}_{\mathrm{1}} \:,\:\:\boldsymbol{\mathrm{S}}_{\mathrm{2}} \:,\:\:\mathrm{S}_{\mathrm{3}} \:\:… \\ $$$$\mathrm{if}\:\mathrm{it}\:\mathrm{correspond} \\ $$
Commented by perlman last updated on 29/May/19
sir ther is mistak in my  answer i will fix it
$${sir}\:{ther}\:{is}\:{mistak}\:{in}\:{my}\:\:{answer}\:{i}\:{will}\:{fix}\:{it} \\ $$$$ \\ $$
Commented by perlman last updated on 29/May/19
the sum is Σ_(k=1) ^(+∞) (1/(3k(3k−1)(3k−2))).and notΣ(1/(k(k+1)(k+2)))sorry for this/mistak
$${the}\:{sum}\:{is}\:\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}\left(\mathrm{3}{k}−\mathrm{1}\right)\left(\mathrm{3}{k}−\mathrm{2}\right)}.{and}\:{not}\Sigma\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}{sorry}\:{for}\:{this}/{mistak} \\ $$
Commented by Tawa1 last updated on 29/May/19
Okay sir, when you check. Tell me sum of the first n terms
$$\mathrm{Okay}\:\mathrm{sir},\:\mathrm{when}\:\mathrm{you}\:\mathrm{check}.\:\mathrm{Tell}\:\mathrm{me}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{n}\:\mathrm{terms} \\ $$
Commented by tanmay last updated on 29/May/19
T_n =(1/([1+(n−1)3][2+(n−1)3][3+(n−1)3]))  T_n =(1/((3n−2)((3n−1)(3n)))  S_n =C−(1/(3×1×(3n−1)3n))  S_1 =C−(1/(18))=T_1 =(1/6)  C=(1/6)+(1/(18))=((3+1)/(18))=(2/9)  S_n =(2/9)−(1/(9n(3n−1)))  S_∞ =(2/9)  pls check...
$${T}_{{n}} =\frac{\mathrm{1}}{\left[\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{3}\right]\left[\mathrm{2}+\left({n}−\mathrm{1}\right)\mathrm{3}\right]\left[\mathrm{3}+\left({n}−\mathrm{1}\right)\mathrm{3}\right]} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{2}\right)\left(\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)\right.} \\ $$$${S}_{{n}} ={C}−\frac{\mathrm{1}}{\mathrm{3}×\mathrm{1}×\left(\mathrm{3}{n}−\mathrm{1}\right)\mathrm{3}{n}} \\ $$$${S}_{\mathrm{1}} ={C}−\frac{\mathrm{1}}{\mathrm{18}}={T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{18}}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{18}}=\frac{\mathrm{2}}{\mathrm{9}} \\ $$$${S}_{{n}} =\frac{\mathrm{2}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{9}{n}\left(\mathrm{3}{n}−\mathrm{1}\right)} \\ $$$${S}_{\infty} =\frac{\mathrm{2}}{\mathrm{9}} \\ $$$${pls}\:{check}… \\ $$

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