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for-each-n-N-f-n-x-nx-1-x-2-n-for-each-x-0-x-1-and-a-n-0-1-f-n-x-dx-if-s-n-sin-pia-n-for-each-n-N-then-li-n-m-s-n-




Question Number 11089 by suci last updated on 11/Mar/17
for each n∈N, f_n (x)=nx(1−x^2 )^n   for each x, 0≤x≤1 and a_n =∫_0 ^1 f_n (x)dx  if s_n =sin(πa_n ), for each n∈N, then  li_(n→∼) m s_n =....???
foreachnN,fn(x)=nx(1x2)nforeachx,0x1andan=01fn(x)dxifsn=sin(πan),foreachnN,thenlimn→∼sn=.???
Commented by FilupS last updated on 11/Mar/17
n∈N  f_n (x)=nx(1−x^2 )^n          0≤x≤1  a_n =∫_0 ^( 1) f_n (x)dx  s_n =sin(πa_n )  lim_(n→∞)  s_n =L  L=?     a_n =∫_0 ^( 1) f_n (x)dx  a_n =∫_0 ^( 1) nx(1−x^2 )^n dx  =n∫_0 ^( 1) x(1−x^2 )^n dx  u=1−x^2   du=−2xdx  =−n(1/2)∫_0 ^( 1) u^n du  =−n(1/2)((1/(n+1))u^(n+1) )_0 ^1   =−(n/(2(n+1)))(u^(n+1) )_0 ^1   =−(n/(2(n+1)))((1−x^2 )^(n+1) )_0 ^1   =−(n/(2(n+1)))(0^(n+1) −1^(n+1) )  =(n/(2(n+1)))  ∴a_n =(n/(2(n+1)))     s_n =sin(πa_n )  s_n =sin(π(n/(2(n+1))))  s_n =sin((1/2)π(n/((n+1))))     lim_(n→∞)  s_n =lim_(n→∞) sin((1/2)π(n/((n+1))))  =sin((1/2)π[lim_(n→∞) (n/((n+1)))])  L′Ho^� pital′s Law  =sin((1/2)π lim_(n→∞)  (1/1))  =sin((1/2)π)  ∴ L = 1     lim_(n→∞)  s_n =1
nNfn(x)=nx(1x2)n0x1an=01fn(x)dxsn=sin(πan)limnsn=LL=?an=01fn(x)dxan=01nx(1x2)ndx=n01x(1x2)ndxu=1x2du=2xdx=n1201undu=n12(1n+1un+1)01=n2(n+1)(un+1)01=n2(n+1)((1x2)n+1)01=n2(n+1)(0n+11n+1)=n2(n+1)an=n2(n+1)sn=sin(πan)sn=sin(πn2(n+1))sn=sin(12πn(n+1))limnsn=limsinn(12πn(n+1))=sin(12π[limnn(n+1)])LHopital^sLaw=sin(12πlimn11)=sin(12π)L=1limnsn=1

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