Rate of convergence of $\left[ \left( \sum\limits_{i=j}^n {2i+1}\right)^{\frac{1}{2}}\right]$

Question

I have $$a_{n} = \left[ \left( \sum\limits_{i=j}^n {2i+1}\right)^{\frac{1}{2}}\right]$$ for some fixed $j\geq1$, where the square brackets are the fractional part. Now I know $$\lim_{n \to \infty} a_{n} = 1$$ as for $n\geq \frac{j^{2}}{2}:(n+1)^2-c_1=b_n\ge n^2+1$ for $c_1=j^2$, where $b_n$ is the sum. But now how fast is $a_n$ converging to $1$? Is the laurent series as good as you can get? ($1-c/(2n)+c/(2n^2)-...$)

Answer 1

For every $n\geqslant\frac12j^2$, $$\sum_{i=j+1}^n(2i+1)=\sum_{i=j+1}^n(i+1)^2-i^2=(n+1)^2-j^2\in[n^2+1,(n+1)^2-1],$$ hence $$a_n=\left\{\sqrt{(n+1)^2-j^2}\right\}=\sqrt{(n+1)^2-j^2}-n=\frac{(n+1)^2-j^2-n^2}{\sqrt{(n+1)^2-j^2}+n},$$ that is, $$a_n=2\frac{1-\frac{j^2-1}{2n}}{1+\sqrt{1+\frac2n-\frac{j^2-1}{n^2}}}=1-\frac{j^2}{2n}+O\left(\frac1{n^2}\right).$$