Method of differences with partial fractionsAQA A-Level Further Maths: Revision notes

A series can be summed in closed form if its general term can be written as a difference $u_r=f(r)-f(r+1)$. When the terms are added, the middle terms cancel in pairs, which is called telescoping: $\sum_{r=1}^{n}\left[f(r)-f(r+1)\right]=f(1)-f(n+1).$ Only the first and last terms survive. Write out the first two or three terms and the last two terms explicitly so that you can see exactly what cancels. Example: $u_r=r^3-(r-1)^3=3r^2-3r+1$, so $\sum_{r=1}^{n}(3r^2-3r+1)=n^3-0^3=n^3$. Combined with $\sum r=\frac{n(n+1)}{2}$ this gives a route to $\sum r^2$. If $u_r=f(r+1)-f(r)$ the answer is $f(n+1)-f(1)$ instead, so check the sign.

Rational terms such as $\frac{1}{r(r+1)}$ do not look like differences, but partial fractions turn them into one: $\frac{1}{r(r+1)}=\frac{1}{r}-\frac{1}{r+1}.$ Find the constants by writing $1\equiv A(r+1)+Br$ and substituting convenient values ($r=0$, $r=-1$), or by equating coefficients. Then $\sum_{r=1}^{n}\frac{1}{r(r+1)}=1-\frac{1}{n+1}=\frac{n}{n+1}$. For $\frac{1}{(2r-1)(2r+1)}=\frac12\left(\frac{1}{2r-1}-\frac{1}{2r+1}\right)$ the sum is $\frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}$. Keep the factor $\frac12$ outside the whole bracket. Always check with $n=1$: the sum must equal the first term.

When the denominators differ by $2$, the cancelling terms are two apart, so two terms survive at each end. For $u_r=\frac{1}{r(r+2)}=\frac12\left(\frac1r-\frac1{r+2}\right)$: $S_n=\frac12\left[1+\frac12-\frac{1}{n+1}-\frac{1}{n+2}\right]=\frac34-\frac{2n+3}{2(n+1)(n+2)}.$ In general, for $f(r)-f(r+k)$ the first $k$ terms of $f$ and the last $k$ terms of $f$ (with the sign reversed) remain. Combine surviving fractions over a common denominator to simplify, and check the result with $n=1$: here $\frac34-\frac{5}{12}=\frac13=u_1$.

A closed form for $S_n$ gives the sum to infinity by letting $n\to\infty$: terms such as $\frac{1}{n+1}$ tend to $0$, so $\sum_{r=1}^{\infty}\frac{1}{r(r+1)}=1$. The series is then convergent. To sum from $r=a$ to $b$ use $S_b-S_{a-1}$. For example $\sum_{r=11}^{20}(3r^2-3r+1)=20^3-10^3=7000$, and $\sum_{r=11}^{\infty}\frac{1}{r(r+2)}=\frac34-S_{10}=\frac{23}{264}$. You may also be asked for the least $n$ such that $S_n$ is within a given tolerance of its limit: set up the inequality in $n$, solve, and round up to a whole number, checking either side.