Compute limn→∞∑k=1narcsin⁡(kn2)

Alisha Pitts

Alisha Pitts

Answered

2022-01-24

Compute
limnk=1narcsin(kn2)

Answer & Explanation

Mazzuranavf

Mazzuranavf

Expert

2022-01-25Added 10 answers

Recall for any θ(0,π2), we have the inequality
sinθ<θ<tanθ
This means for any x(0,1), we have the bound
x<arcsinx<x1x2
For any n>1, this leads to
n+12n=k=1nkn2<k=1narcsinkn2
<k=1nkn21(kn2)2111n2k=1nkn2
=n+12n21
As n, it is clear both sides converge to 12. By squeezing, we obtain
limnk=1narcsinkn2=12
ebbonxah

ebbonxah

Expert

2022-01-26Added 15 answers

Another way to look at this is to observe that
arcsin(kn2)=kn201du1k2n4u2
Then you can reverse order of summation and integration and get that the sum equals
01du1nk=1n(kn)1k2n4u2
We almost have a Riemann sum, but not quite. The good news is that we can convert this to a Riemann sum by subbing u=nv in the integral. The result is
n01ndv1nk=1n(kn)1k2n2v2
Now we have a Riemann sum, and as n it becomes the integral
01dxx1v2x2=11v2v2
The limit we seek is then
limn(n01ndv11v2v2)=limn(n0arcsin1ndθcosθ1+cosθ)
which 1/2.
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

Here is a nonrigorous approach that can be filled out to a solution: Observe that each kn2 is going to be pretty small once n is large. For small angles θ you have that sinθθ so likewise arcsinxx for small x. This means that k=1narcsin(kn2)k=1nkn2=1n2k=1nk =n(n+1)2n2

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