How to solve this sum of fractions in terms of $k$?

Question

I'm interested in solving the series $$\frac{2}{1\cdot3}+\frac{3}{3\cdot5}+\frac{4}{5\cdot7}\cdots+\frac{n+1}{(2n-1)(2n+1)}.$$

That is, $$\sum^n_{k=1}\frac{k+1}{(2k-1)(2k+1)}.$$

Can we get a concise representation for this summation?

Answer 1

Hint :

Partial fraction expansion

$\frac{3}{4(2k-1)} - \frac {1}{4(2k+1)}$

Answer 2

Using partial fraction decomposition$$S_n=\sum^n_{k=1}\frac{k+1}{(2k-1)(2k+1)}=\frac34\sum^n_{k=1}\frac{1}{ (2 k-1)}-\frac14\sum^n_{k=1}\frac{1}{ (2 k+1)}$$ $$S_n=\frac{1}{4}H_{n-\frac{1}{2}}+\frac{ n (1+2\log (2))+\log (2)}{(2 n+1)}$$

If $n$ is large, using the asymptotics of harmonic numbers $$S_n=\frac{1}{4} \left(\gamma +1+2\log (2)\right)+\frac{1}{4}\log(n)-\frac{1}{8 n}+\frac{7}{96 n^2}-\frac{1}{32 n^3}+O\left(\frac{1}{n^4}\right)$$

Try for $n=10$; the exact value is $$S_{10}=\frac{18983113}{14549535}\approx 1.3047230$$ while the above truncated expansion would give $1.3047217$.