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Question Number 128846 by Ar Brandon last updated on 10/Jan/21

u_n =Σ_(k=1) ^n sin(((kπ)/n))sin((k/n^2 ))  Find lim_(n→∞) u_n

un=nk=1sin(kπn)sin(kn2)Double subscripts: use braces to clarify

Commented by Dwaipayan Shikari last updated on 10/Jan/21

lim_(n→∞) sin((k/n^2 ))=(k/n^2 )  u_n =lim_(n→∞) (1/n)Σ_(k=1) ^n (k/n)sin((kπ)/n)  =∫_0 ^1 xsinxπ dx = (1/π^2 )∫_0 ^π u sinu =(1/π^2 )[−u cosu]_0 ^π +(1/π^2 )∫_0 ^π cosu du  =(1/π)

limnsin(kn2)=kn2un=limn1nnk=1knsinkπn=01xsinxπdx=1π20πusinu=1π2[ucosu]0π+1π20πcosudu=1π

Commented by Dwaipayan Shikari last updated on 10/Jan/21

(1/π)=((2(√2))/(9801))Σ_(k=0) ^∞ (((4k)!(1103+26390k))/((k!)^4 396^(4k) ))

1π=229801k=0(4k)!(1103+26390k)(k!)43964k

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