Question Number 100297 by Coronavirus last updated on 26/Jun/20

calulate using Riemann sums  tbe limit of this sequence    Σ_(k=n) ^(2n) sin ((π/k))

Answered by abdomsup last updated on 26/Jun/20

A_n =_(p=k−n)   Σ_(p=0) ^n  sin((π/(p+n)))  we have x−(x^3 /6)≤sinx≤x ⇒  (π/k)−(π^3 /(6k^3 ))≤sin((π/k))≤(π/k)  ⇒Σ_(k=n) ^(2n)  (π/k)−π^3  Σ_(k=n) ^(2n)  (1/k^3 )  ≤Σ_(k=n) ^(2n)  sin((π/k))≤πΣ_(k=n) ^(2n)  (1/k)  π Σ_(p=0) ^n  (1/(p+n))−π^(3 ) Σ_(p=0) ^n  (1/((p+n)^3 ))  ≤Σ_(k=n) ^(2n)  sin((π/k))≤πΣ_(p=0) ^n  (1/(p+n))  we haveΣ_(p=0) ^n  (1/(p+n))  (1/n)+(1/(n+1))+....+(1/(2n))  1+(1/n)+....+(1/(n−1))+....+(1/(2n))  −H_(n−1) =H_(2n) −H_(n−1)   =ln(2n)+γ +0((1/(2n)))−ln(n−1)  −γ−o((1/(n−1)))→ln2  its clear  that Σ_(p=0) ^n  (1/((p+n)^3 ))→0  because Σ)...)≤Σ_0 ^n  (1/n^3 )=((n+1)/n^3 )  ⇒lim_(n→+∞) Σ_(k=n) ^(2n)  sin((k/n))=πln2

Commented byCoronavirus last updated on 26/Jun/20

Thanks you Mr

Commented byDGmichael last updated on 26/Jun/20

😀😀😀

Commented bymathmax by abdo last updated on 26/Jun/20

you are welcome sir

Commented byAr Brandon last updated on 26/Jun/20

Qui a compris ? 😅😅

Commented byAr Brandon last updated on 26/Jun/20

Sir is H_n  a special function ?

Commented bymathmax by abdo last updated on 26/Jun/20

H_n =1+(1/2)+(1/3) +...+(1/n) is harmonic sequence

Commented byAr Brandon last updated on 27/Jun/20

OK thank you Sir