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n-1-m-log-n-n-3-2-




Question Number 63139 by Tawa1 last updated on 29/Jun/19
Σ_(n = 1) ^m  ((log n)/n^(3/2) )
mn=1lognn3/2
Commented by Tawa1 last updated on 30/Jun/19
please help with this sir   Test for convergence.     Σ_(n = 1) ^∞   (1/(n^3  sin^2 n))
pleasehelpwiththissirTestforconvergence.n=11n3sin2n
Commented by Tawa1 last updated on 30/Jun/19
God bless you sir
Godblessyousir
Commented by mathmax by abdo last updated on 30/Jun/19
if you need convergence  let S_m =Σ_(n=1) ^m  ((ln(n))/n^(3/2) )  we have lim_(n→+∞)  S_m =Σ_(n=2) ^∞   ((ln(n))/n^(3/2) )   let ϕ(x) =((ln(x))/x^(3/2) )  with x>1  we have ϕ^′ (x) =(((x^(3/2) /x)−ln(x)(3/2)x^(1/2) )/x^3 ) =(((√x)−(3/2)(√x)ln(x))/x^3 ) =((√x)/x^3 ){1−(3/2)ln(x)}  ⇒∃n_0   >0/ for x>n_0    ϕ is decreazing on]x_0  ,+∞[ so  the serie and   ∫_2 ^(+∞)  ((ln(x))/x^(3/2) ) dx have the same nature but  ∫_2 ^(+∞)   ((ln(x))/x^(3/2) )dx =_(ln(x)=t)      ∫_(ln(2)) ^(+∞)     (t/e^((3/2)t) ) e^(t ) dt = ∫_(ln(2)) ^(+∞)    t  e^((1−(3/2))t) dt  =∫_(ln(2)) ^(+∞)   te^(−(t/2))  dt     this integral converges because lim_(t→+∞)  t^2 (t e^(−(t/2)) )=0 so  Σ_(n=1) ^∞  ((ln(n))/n^(3/2) )  converges .
ifyouneedconvergenceletSm=n=1mln(n)n32wehavelimn+Sm=n=2ln(n)n32letφ(x)=ln(x)x32withx>1wehaveφ(x)=x32xln(x)32x12x3=x32xln(x)x3=xx3{132ln(x)}n0>0/forx>n0φisdecreazingon]x0,+[sotheserieand2+ln(x)x32dxhavethesamenaturebut2+ln(x)x32dx=ln(x)=tln(2)+te32tetdt=ln(2)+te(132)tdt=ln(2)+tet2dtthisintegralconvergesbecauselimt+t2(tet2)=0son=1ln(n)n32converges.

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