Kendrick Hampton

Answered

2022-06-25

I would like to find an equivalent of
${u}_{n}-{u}_{\mathrm{\infty }}=\sum _{k=1}^{n}\frac{n}{{n}^{2}+{k}^{2}}-{u}_{\mathrm{\infty }}$

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Recalculate according to your conditions!

Answer & Explanation

laure6237ma

Expert

2022-06-26Added 27 answers

Your numerical work indeed leads to the right conjecture ${u}_{n}-\frac{\pi }{4}\sim -\frac{1}{4n}$. I am feeling lazy, so to prove the result I will appeal to a standard result about $\text{TRAP}\left(n\right)$, the Trapezoidal Rule with division into n equal parts. It is known that under suitable differentiability assumptions, which are amply met here, the error in $\text{TRAP}\left(n\right)$ is $O\left(1/{n}^{2}\right)$. Note that
$\text{TRAP}\left(n\right)=\sum _{k=1}^{n-1}\frac{n}{{n}^{2}+{k}^{2}}+\frac{1}{2}\left(\frac{n}{{n}^{2}}+\frac{n}{2{n}^{2}}\right).$
Thus
$\text{TRAP}\left(n\right)=\sum _{k=1}^{n}\frac{n}{{n}^{2}+{k}^{2}}+\frac{1}{2}\left(\frac{n}{{n}^{2}}+\frac{n}{2{n}^{2}}\right)-\frac{n}{2{n}^{2}}=\sum _{k=1}^{n}\frac{n}{{n}^{2}+{k}^{2}}+\frac{1}{4n}.$
It follows that
$\sum _{k=1}^{n}\frac{n}{{n}^{2}+{k}^{2}}=\frac{\pi }{4}-\frac{1}{4n}+O\left(1/{n}^{2}\right).$

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