Compute $\sum_{k=1}^{25} (\frac{1}{k}-\frac{1}{k+4})$ 
Compute $\sum_{k=1}^{25} (\frac{1}{k}-\frac{1}{k+4})$

I know that some of the terms will cancel each other. Have it been $k+1$ instead of $k+4$, I could have easily see the pattern in which the terms cancel each other. I don't know what would I do if it was say $k+6$ or bigger.
How can I identify the pattern in these type of problems?
 A: You may write
$$\frac{1}{k} - \frac{1}{k+4} = \left( \frac{1}{k} - \frac{1}{k+1} \right) + \left( \frac{1}{k+1} - \frac{1}{k+2} \right)+ \left( \frac{1}{k+2} - \frac{1}{k+3} \right) + \left( \frac{1}{k+3} - \frac{1}{k+4} \right)$$
Now, apply telescoping and you get for the sum
$$\left( \frac{1}{1} - \frac{1}{26} \right) + \left( \frac{1}{2} - \frac{1}{27} \right)+ \left( \frac{1}{3} - \frac{1}{28} \right) + \left( \frac{1}{4} - \frac{1}{29} \right)$$
A: Note that for $0\leq b\leq N$,
$$\sum_{k=1}^{N} (\frac{1}{k}-\frac{1}{k+b})=
\sum_{k=1}^{N} \frac{1}{k}-\sum_{k=1}^{N}\frac{1}{k+b}=\sum_{k=1}^{N} \frac{1}{k}-\sum_{k=1+b}^{N+b}\frac{1}{k}.$$
Compare the terms in the two sums, what do we obtain after deleting the opposite terms?
A: This may help you:
\begin{align}
\require{cancel}
\sum\limits_{k=1}^N \left(\frac{1}{k} - \frac{1}{k+4}\right) &= \sum\limits_{k=1}^N \frac1k - \sum\limits_{k=1}^N \frac{1}{k+4}\\ &= \sum\limits_{k=1}^N \frac1k - \sum\limits_{k=1+4}^{N+4}\frac1k \\&=
\left(\sum\limits_{k=1}^4\frac1k + \sum\limits_{k=5}^N\frac1k\right) - \left(\sum\limits_{k=5}^N\frac1k + \sum\limits_{k=N+1}^{N+4}\frac1k\right)\\ &=
\left(\sum\limits_{k=1}^4\frac1k + \cancel{\sum\limits_{k=5}^N\frac1k}\right) - \left(\cancel{\sum\limits_{k=5}^N\frac1k} + \sum\limits_{k=N+1}^{N+4}\frac1k\right)\\ &=
\sum\limits_{k=1}^4 \frac1k - \sum\limits_{k=N+1}^{N+4}\frac1k
\end{align}
First split up the sum into two sums. Then change index so that the summand is the same in each. Then realize that a lot of the terms cancel out. Lastly, you're only left with a few terms.
A: The general term is $\frac 1k -\frac{1}{k+4}$.
The first right term is $\frac 15$, so $1, \frac 12,\frac13,\frac14 $ will not be cancelled on the left.
The last left term is $\frac{1}{25}$, so the terms $\frac{1}{26}, \frac{1}{27}, \frac{1}{28}, \frac{1}{29}$ will not be cancelled on the right. Collate these results.
