Summation by Parts

Your computations are correct, which is easy to see by differencing.

The anti-difference of $x_{-1} = \frac{1}{x+1}$ is known a Harmonic number: $$ \sum_x \frac{1}{x+1} = H_x $$ where $H_n = \sum_{k=1}^n \frac{1}{k}$.

I'm not sure this fully answers your questions, but it is easier to solve this problem without summation by part, simply by partial fraction expansion: $$ S_n = \sum_{k=1}^{n}\frac{1}{k} + \sum_{k=1}^{n}\frac{1}{k+1}=2H_n-1+\frac{1}{n+1}=2H_n-\frac{n}{n+1} $$ where $H_n$ is n'th harmonic number