Evaluate $\sum\limits_{r=1}^{50}\left[\frac{1}{49+r} - \frac{1}{2r(2r-1)}\right]$

Question

Find $$ S=\sum_{r=1}^{50}\left[\frac{1}{49+r} - \frac{1}{2r(2r-1)}\right] $$ Thus, some terms are in harmonic progression.

I tried to rearrange $S$ and represent it as a sum of two terms.

Answer 1

$S=\sum\limits_{r=1}^{50}\left[\frac{1}{49+r} - \frac{1}{2r(2r-1)}\right]$

The first part of the sum:

$\sum\limits_{r=1}^{50}\frac{1}{49+r}=\sum\limits_{r=50}^{99}\frac{1}{r}=\sum\limits_{r=1}^{99}\frac{1}{r}-\sum\limits_{r=1}^{49}\frac{1}{r}$

The second part of the sum:

$\sum\limits_{r=1}^{50} - \frac{1}{2r(2r-1)}=\sum\limits_{r=1}^{50}\big( \frac{1}{2r}{-\frac{1}{(2r-1)}}\big)=\frac{1}{100}-\sum\limits_{r=1}^{99}\frac{(-1)^{r-1}}{r}$

Finally

$S=\frac{1}{100}+\sum\limits_{r=1}^{99}\big(\frac{1}{r}-\frac{(-1)^{r-1}}{r}\big)-\sum\limits_{r=1}^{49}\frac{1}{r}$ as the 99. term is equal to zero in the first part of the sum and $r=2k$ we get the followings:

$S=\frac{1}{100}+\sum\limits_{k=1}^{49}\frac{2}{2k}-\sum\limits_{r=1}^{49}\frac{1}{r}=\frac{1}{100}$