Evaluate $\sum \limits_{n = 1}^{\infty} \ln (2n) - \frac{\ln (2n + 1) + \ln (2n - 1)}{2}$
I have no idea how I should approach this problem.
So far, I've only studied partial sums and Riemann integrals, but I don't really understand how and when apply these concepts.
It seems like I can't calculate this, using partial sums, since there is no explicit formula for $S_n$.
On
That is clearly related with Wallis' product, but if we lack such knowledge, we may simply exploit the integral representation for the logarithm provided by Frullani's integral: $$ \forall a,b>0,\qquad \int_{0}^{+\infty}\frac{e^{-ax}-e^{-bx}}{x}\,dx = \log\frac{b}{a}\tag{1}$$ allowing us to write the given series (that is absolutely convergent, since its general term behaves like $\frac{1}{n^2}$) as: $$ \frac{1}{2}\sum_{n\geq 1}\int_{0}^{+\infty}\frac{e^{-(2n+1)x}+e^{-(2n-1)x}-2e^{-2nx}}{x}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{1-e^{-x}}{1+e^x}\cdot\frac{dx}{x} \tag{2}$$ The problem now boils down to showing that $$ \int_{0}^{+\infty}\frac{\tanh x}{x}e^{-2x}\,dx = \log\frac{\pi}{2} \tag{3}$$ that is an instance of the more general $$ \int_{0}^{+\infty}\frac{\tanh x}{x}e^{-mx}\,dx = \log\,\left(\frac{4\,\Gamma\left(\frac{m+4}{4}\right)^2}{m\,\Gamma\left(\frac{m+2}{4}\right)^2}\right).\tag{4}$$
The classic result due to Wallis is \begin{eqnarray*} \prod_{n=1}^{\infty} \frac{2n} {2n-1} \frac{2n} {2n+1} =\frac{\pi} {2} \end{eqnarray*} So your sum is $\frac{1}{2} \ln (\frac{\pi} {2})$.