Evaluate $\sum_{n=0}^\infty\left(-1+(n+1)\ln\left(\frac{2n+3}{2n+1}\right)\right)$ I proved that the given series is convergent and to find the sum I tried to compute the partial sum and then I pass to the limit but I didn't find a closed form to this partial sum. How I should proceed?
 A: Set $a_n=n\ln(2n+1)$, and rewrite the general term of the series as \begin{align*}b_n&=-1+(n+1)\ln\left(\frac{2n+3}{2n+1}\right)\\&=-1+(n+1)\ln(2n+3)-n\ln(2n+1)-\ln(2n+1)=-1-\ln(2n+1)+a_{n+1}-a_n.\end{align*} Summing from $k=0$ to $n$, we obtain that $$\sum_{k=0}^nb_k=\sum_{k=0}^n\left(-1-\ln(2k+1)+a_{k+1}-a_k\right)=-(n+1)-\sum_{k=0}^n\ln(2k+1)+a_{n+1},$$ since $a_0=0$, therefore \begin{align*}\sum_{k=0}^nb_k&=-(n+1)-\sum_{k=0}^n\ln(2k+1)+(n+1)\ln(2n+3)=\ln\left(\frac{(2n+3)^{n+1}e^{-n-1}}{3\cdot 5\cdot 7\cdots\cdot (2n+1)}\right)\\
&=\ln\left(\frac{(2n+3)^{n+1}e^{-n-1}\cdot 2^nn!}{(2n+1)!}\right).\end{align*} For this last term, use Stirling's formula: asymptotically, it is equal to 
$$\ln\left(\frac{(2n+3)^{n+1}e^{-n-1}\cdot 2^nn^ne^{-n}\sqrt{2\pi n}}{(2n+1)^{2n+1}e^{-2n-1}\sqrt{2\pi(2n+1)}}\right)\simeq\ln\left(\frac{(2n+3)^{n+1}(2n)^n}{(2n+1)^{2n+1}\sqrt{2}}\right),$$ and $$
\frac{(2n+3)^{n+1}(2n)^n}{(2n+1)^{2n+1}\sqrt{2}}=\left(1+\frac{2}{2n+1}\right)^{n+1}\left(1+\frac{1}{2n}\right)^{-n}\frac{1}{\sqrt{2}}\xrightarrow[n\to\infty]{}\frac{e\cdot e^{-1/2}}{\sqrt{2}}=\frac{e^{1/2}}{\sqrt{2}},$$ therefore the series is equal to $$\ln\left(\frac{e^{1/2}}{\sqrt{2}}\right)=\frac{1}{2}-\frac{1}{2}\ln 2.$$
