Sunday, 22 January 2017

real analysis - Compute limntoinftyintfracpi20sumnk=1(sinx)kdx




Compute limnπ20nk=1(sinx)kdx.





I tried writing limnπ20nk=1(sinx)kdx=limnπ201(sinx)n+11sinxdx, but I don't know how to continue form here.



I know that the limit diverges to , but I don't know how to prove it.


Answer



The wanted limit is +. Indeed, by letting Ik=π/20(sinx)kdx we have that {Ik}k1 is a decreasing sequence, but it is also log-convex by the Cauchy-Schwarz inequality, and by integration by parts
I2kIkIk+1=π2(k+1)


such that
nk=1Ikπ2nk=11k+1π2nk=12(k+2k+1)=2π(n+22).


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