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

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.

• So what is $HP$, and why wouldn't you spell it out anyway? Commented May 3, 2018 at 17:41
• @DavidG.Stork I believe it's harmonic progression. Commented May 3, 2018 at 18:07
• The answer is $$\frac{1}{100}$$ by the way. Still figuring out the working... Commented May 3, 2018 at 18:21
• You need a $\sum$ symbol on the second one$\ldots$ Commented May 3, 2018 at 18:39
• $New$ user info: Please, you must always show what you tried. That enhances the possibility that MSE-people pay attention to your post. In addition, read the MSE-$\LaTeX$-$\texttt{MathJax}$ Tutorial. Commented May 3, 2018 at 18:45

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

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:

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