What tools would you gladly recommend me for computing precisely

ardeinduie

ardeinduie

Answered question

2022-01-23

What tools would you gladly recommend me for computing precisely the limit below? Maybe a starting point?
limn1log(n)k=1ncos(sin(2πlog(k)))k

Answer & Explanation

trovabile4p

trovabile4p

Beginner2022-01-24Added 13 answers

Let f(x)=cos(sin(2πx))x, we have
f(x)=2πcos(2πx)sin(sin(2πx))+cos(sin(2πx))x2|f(x)|2π+1x2
By MVT, for any x(k,k+1] we can find a ζ(0,1) such that
|f(x)f(k)|=|f(k+ζ(xk))|(xk)2π+1k2
This implies
|kk+1f(x)dxf(k)|2π+1k2
As a result,
|k=1nf(k)1nf(x)dx||f(n)|+k=1n1|f(k)kk+1f(x)dx|1+(2π+1)k=11k2
=1+(2π+1)π26<
As a result,
limn1lognk=1nf(k)=limn1logn1nf(x)dx
=limL1L0Lcos(sin(2πt))dt
Damian Roberts

Damian Roberts

Beginner2022-01-25Added 14 answers

I would start with this:
0|1log(n)k=0ncos(sin(2πlog(k)))k|1lognk=0n1k
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Using the standard estimate for the harmonic numbers, |1lognk=1ncos(sin(2πlog(k)))k|1lognk=1n1k =1logn(log(n)+γ+O(1/n)) So in the limit, |limn1lognk=1ncos(sin(2πlog(k)))k|1 This is a pretty elementary bound, so you were probably already aware of it, but hopefully it helps.

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