Series representation 
I have started this problem by expanding it so that i can get some cancellation term, but couldn't reach on the correct result.I got the result like -ln4 -ln5-ln6.....  Please have a look on this.

Consider the sum:
$$S=\sum_{n=1}^m \left(1-\frac{2n+5}{2}\ln\left(\frac{n+3}{n+2}\right)\right)=m-\frac{1}{2}\sum_{n=1}^m \ln\left(\frac{n+3}{n+2}\right)^{2n+5}$$
Since,
$$\sum_{n=1}^m \ln\left(\frac{n+3}{n+2}\right)^{2n+5}=\ln\left(\frac{4^7}{3^7}\cdot \frac{5^9}{4^9}\cdot \frac{6^{11}}{5^{11}}\cdots\frac{(m+3)^{2m+5}}{(m+2)^{2m+5}}\right)=\ln\left(\frac{4(m+3)^{2m+5}}{3^5((m+2)!)^2}\right)$$
Hence,
$$S=\ln(e^m)-\ln\left(\frac{2(m+3)^{\frac{2m+5}{2}}}{3^{5/2}(m+2)!}\right)=\ln\left(\frac{3^{5/2}}{2}\cdot \frac{e^m(m+2)!}{(m+3)^{\frac{2m+5}{2}}}\right)$$
Calculate the limit of the final expression as $m\rightarrow \infty$ using Stirling's approximation to obtain: 
$$S=\ln\left(\frac{3^{5/2}}{2}\cdot \frac{\sqrt{2\pi}}{e^3}\right)=\boxed{\dfrac{1}{2}\ln(6\pi)+\ln\left(\dfrac{9}{2}\right)-3}$$
