I am trying to solve $\sum\limits\_{k=1}^n\frac{2k+1}{k(k+1)}$ using summation by parts:

$\sum u\Delta v=uv-\sum Ev\Delta u$

$u = 2x+1, \Delta v=1/x(x+1)=(x-1)\_{-2}$

$v=-(x-1)\_{-1}=-1/x$

$\Delta u = 2, Ev = -1/(x+1)$

$\sum\frac{(2x+1)}{x(x+1)}\delta x=\frac{-2x-1}{x}+2\left(\sum x\_{-1}\delta x\right)$

Is my solution correct up to this point? If it is, what is the anti-difference $\sum x\_{-1}\delta x$? This seems to be the finite equivalent to $\int x^{-1}dx=ln(x)$.