I love being inspired! (Part-2)

$\displaystyle \sum_{n=1}^{2015} \frac{n^{\log_\frac{n+1}{n}n}} {(n+1)^{\log_\frac{n+1}{n}n+1}}$

Let's simplify the above expression ,

Let $x=\log_\frac{n+1}{n}n$ and $y=\log_\frac{n+1}{n}n+1$ .

Now ,

$y=\log_\frac{n+1}{n}(\frac{n+1}{n}×n)=\log_\frac{n+1}{n}\frac{n+1}{n}+\log_\frac{n+1}{n}n$

Therefore , $y=1+x$ .

$\rightarrow \frac{n^{\log_\frac{n+1}{n}n}} {(n+1)^{\log_\frac{n+1}{n}n+1}}=\frac{n^{x}}{(n+1)^{y}}=\frac{n^{x}}{(n+1)^{x+1}}$

$\rightarrow \frac{n^{x}}{(n+1)^{x}}×\frac{1}{n+1}=(\frac{n+1}{n})^{-x}×\frac{1}{n+1}$

$\rightarrow (\frac{n+1}{n})^{-\log_\frac{n+1}{n}n}×\frac{1}{n+1}$

$\rightarrow \frac{1}{n}×\frac{1}{n+1}=\frac{1}{n(n+1)}$

Now , converting $\frac{1}{n(n+1)}\rightarrow$ Partial fraction .

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

Therefore ,

$x=\displaystyle \sum_{n=1}^{2015} \frac{n^{\log_\frac{n+1}{n}n}} {(n+1)^{\log_\frac{n+1}{n}n+1}}=\displaystyle \sum^{2015}_{n=1}\frac{1}{n}-\frac{1}{n+1}$

$=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}$

$=1-\frac{1}{2016}=\frac{2015}{2016}$

$2015+2016=\boxed{4031}$

$\begin{aligned} S & = \sum_{n=1}^{2015} \frac{\color{#3D99F6}{n}^{\log_{\frac{n+1}{n}}(n)}}{(n+1)^{\log_{\frac{n +1}{n}}(n+1)}} \\ & = \sum_{n=1}^{2015} \frac{\left(\frac{n+1}{n}\right)^{\log_{\frac{n+1}{n}}(n) \log_{\frac{n+1}{n}}(n)}}{(n+1)^{\log_{\frac{n +1}{n}}(n+1)}} \quad \quad \color{#3D99F6}{n = \left(\frac{n+1}{n}\right)^{\log_{\frac{n+1}{n}}(n)}} \\ & = \sum_{n=1}^{2015} \frac{\left(\frac{n+1}{n}\right)^{\log_{\frac{n+1}{n}}^2(n)}} {\left(\frac{n+1}{n}\right)^{\log_{\frac{n+1}{n}}^2(n+1)}} \\ & = \sum_{n=1}^{2015} \left(\frac{n+1}{n}\right)^{-\left(\log_{\frac{n+1}{n}}^2(n+1) - \log_{\frac{n+1}{n}}^2(n)\right)} \\ & = \sum_{n=1}^{2015} \left(\frac{n+1}{n}\right)^{-\left(\log_{\frac{n+1}{n}}(n+1) + \log_{\frac{n+1}{n}}(n)\right)\left(\log_{\frac{n+1}{n}}(n+1) - \log_{\frac{n+1}{n}}(n)\right)} \\ & = \sum_{n=1}^{2015} \left(\frac{n+1}{n}\right)^{-\left(\log_{\frac{n+1}{n}}(n(n+1)) \right) \left(\color{#3D99F6}{\log_{\frac{n+1}{n}}\left(\frac {n+1}{n}\right)}\right)} \quad \quad \color{#3D99F6}{\log_{\frac{n+1}{n}}\left(\frac {n+1}{n}\right) = 1} \\ & = \sum_{n=1}^{2015} \frac{1}{\left(\frac{n+1}{n}\right)^{\left( \log_{\frac{n+1}{n}}(n(n+1)) \right)}} \\ & = \sum_{n=1}^{2015} \frac{1}{n(n+1)} = \sum_{n=1}^{2015} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1 - \frac{1}{2016} = \frac{2015}{2016} \\ \\ \Rightarrow A+B & = \boxed{4031} \end{aligned}$