Finding a closed formula for the sum $\sum_{i=0}^{n-1}\frac{1}{1-(\frac{i}{n})^2}$

Question

Consider the sum: $F(n)=\sum_{i=0}^{n-1}\frac{1}{1-(\frac{i}{n})^2}$.

So far I found these upper and lower bounds using the well known sum-integral inequality:

• $F(n)\geq1+\int_{1}^{n-1}\frac{1}{1-\left(\frac{x}{n}\right)^{2}}\text{ d}x\geq\frac{n}{2}\log(2n-1)-O(\frac{1}{n})$
• $F(n)\leq\frac{1}{1-\left(\frac{n-1}{n}\right)^{2}}+\int_{0}^{n-1}\frac{1}{1-\left(\frac{x}{n}\right)^{2}}\text{ d}x\leq\frac{n}{2}\log(2n-1)+O(n)$

I am trying to find a close formula for $F$, or at least to find a function $G$ such that: $$F(n)=\frac{n}{2}\log(2n-1)+G(n)+o(1)$$

I would guess $G=\Theta(n)$ but it is a pure guess based on looking at graphs...

Thanks!

Answer 1

$$ \sum_{i=0}^{n-1}\frac{n^2}{n^2-i^2}=\sum_{i=0}^{n-1}\frac{n^2}{(n-i)(n+i)}=\frac n2\sum_{i=0}^{n-1}\left(\frac1{n-i}+\frac1{n+i}\right)=\frac n2\left(\sum_{k=1}^{2n-1}\frac1k+\frac1n\right)=\frac{nH_{2n-1}+1}2\;, $$

where $H_n$ is the $n$-th harmonic number with

$$ H_k=\log k+\gamma+\frac1{2k}+O\left(\frac1{k^2}\right)\;, $$

where $\gamma$ is the Euler-Mascheroni constant. Thus

\begin{eqnarray} \sum_{i=0}^{n-1}\frac{n^2}{n^2-i^2} &=& \frac12\left(n\log(2n-1)+n\gamma+\frac n{2(2n-1)}+1+O\left(\frac1n\right)\right) \\ &=&\frac n2\log(2n-1)+\frac{n\gamma}2+\frac58+O\left(\frac1n\right) \\ &=&\frac n2\log(2n)+\frac{n\gamma}2+\frac38+O\left(\frac1n\right) \end{eqnarray}