The method of differencesEdexcel International A Level Further Maths: Revision notes

Some series can be summed by writing each term as a difference $u_r=f(r)-f(r+1)$. When you add the terms from $r=1$ to $n$, almost everything cancels (the series telescopes): $\sum_{r=1}^{n}\big[f(r)-f(r+1)\big]=f(1)-f(n+1).$ Write out the first three and last two terms so that you can see which terms survive. The other pattern, $f(r+1)-f(r)$, gives $f(n+1)-f(1)$.

For a fraction with factors in the denominator, split it first. For example: $\frac{1}{r(r+1)}=\frac1r-\frac1{r+1},\qquad \frac{1}{r(r+2)}=\frac12\left(\frac1r-\frac1{r+2}\right).$ Find the constants by substituting convenient values of $r$ into $1=A(r+2)+Br$ (here $r=0$ and $r=-2$). Then $f(r)=\frac1r$ and the second part is $f(r+1)$ or $f(r+2)$.

Find $\sum_{r=1}^{n}\frac{1}{r(r+1)}$. $\sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right)=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\dots+\left(\frac1n-\frac1{n+1}\right)=1-\frac{1}{n+1}=\frac{n}{n+1}.$ As $n\to\infty$, $\frac{1}{n+1}\to0$, so the series converges to $1$.

If $f(r)=\frac1r$ is paired with $f(r+2)$, then $\frac1{r+2}$ cancels with the $\frac1r$ two terms later, so four terms survive: the first two and the last two. $\sum_{r=1}^{n}\frac12\left(\frac1r-\frac1{r+2}\right)=\frac12\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right).$ For $\frac{1}{(2r-1)(2r+1)}=\frac12\left(\frac{1}{2r-1}-\frac{1}{2r+1}\right)$ the sum is $\frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}$.

The method also works with other functions. Since $(r+1)!-r!=r\cdot r!$: $\sum_{r=1}^{n}r\cdot r!=(n+1)!-1.$ To start the sum at $r=m$ instead of $1$, keep the same $f(r)$ and use $f(m)$ as the first term, for example $\sum_{r=3}^{5}r\cdot r!=6!-3!=714$. For a three-factor fraction, $\frac{1}{r(r+1)(r+2)}=\frac12\left[\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}\right]$, so $f(r)=\frac1{r(r+1)}$. A sum to infinity is found by letting $n\to\infty$ in the result; the error after $n$ terms is the leftover term.