Compute limn→∞∑k=1narcsin⁡(kn2)

Question

Compute  limn→∞∑k=1narcsin⁡(kn2)

Mazzuranavf · Accepted Answer

Recall for any θ∈(0,π2), we have the inequality  sin⁡θ&amp;lt;θ&amp;lt;tan⁡θ  This means for any x∈(0,1), we have the bound  x&amp;lt;arcsin⁡x&amp;lt;x1−x2  For any n&amp;gt;1, this leads to  n+12n=∑k=1nkn2&amp;lt;∑k=1narcsin⁡kn2  &amp;lt;∑k=1nkn21−(kn2)2≤11−1n2∑k=1nkn2  =n+12n2−1  As n→∞, it is clear both sides converge to 12. By squeezing, we obtain  limn→∞∑k=1narcsin⁡kn2=12

ebbonxah · Accepted Answer

Another way to look at this is to observe that  arcsin⁡(kn2)=kn2∫01du1−k2n4u2  Then you can reverse order of summation and integration and get that the sum equals  ∫01du1n∑k=1n(kn)1−k2n4u2  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  n∫01ndv1n∑k=1n(kn)1−k2n2v2  Now we have a Riemann sum, and as n→∞ it becomes the integral  ∫01dxx1−v2x2=1−1−v2v2  The limit we seek is then  limn→∞(n∫01ndv1−1−v2v2)=limn→∞(n∫0arcsin⁡1ndθcos⁡θ1+cos⁡θ)  which 1/2.

RizerMix · Accepted Answer

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 sin⁡θ≈θ so likewise arcsin⁡x≈x for small x. This means that ∑k=1narcsin⁡(kn2)≈∑k=1nkn2=1n2∑k=1nk =n(n+1)2n2

Compute \lim_{n\to\infty}\sum_{k=1}^n\arcsin(\frac{k}{n^2})

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