How do I sum the series $\sum_{n=1}^{\infty} \left( n\cdot \log\frac{2n+1}{2n-1} - 1\right)$

Question

How do I sum the series $$\sum_{n=1}^{\infty} \left( n\cdot \ln\frac{2n+1}{2n-1} - 1\right)$$

My attempts:

• Then $nth$ term can be found to be $-\ln(3 \cdot 5 \cdots (2n-1)) + n\cdot \log(2n+1) -n$.

What to do after this?

Answer 1

You are on the right track. $$S(m) = \sum_{n=1}^m \left(n \ln \left(\dfrac{2n+1}{2n-1}\right) -1\right) = - \ln \left( \dfrac{(2m)!}{2^m m!}\right) + m \ln(2m+1) - m$$ Now use Stirling and finish it off.