Express \[\frac{2}{r^2+4r+3}\] in partial fractions.
First we factorise the denominator as \[\frac{2}{r^2+4r+3}=\frac{2}{(r+1)(r+3)}.\]
Next, it makes sense to guess that there exist numbers \(A\) and \(B\) such that we have the identity \[\begin{equation}\label{eq:partial} \frac{2}{(r+1)(r+3)}=\frac{A}{r+1}+\frac{B}{r+3}. \end{equation}\]Forming a common denominator for the right hand side of this, we find that \[\frac{2}{(r+1)(r+3)}=\frac{A(r+3)+B(r+1)}{(r+1)(r+3)}=\frac{(A+B)r+3A+B}{(r+1)(r+3)}.\] Since the numerator must be equal to \(2\) for all values of \(r\), we obtain the system of equations \[A+B=0, \quad 3A+B=2.\] Subtracting these equations gives \(2A=2\), so \(A=1\) and therefore \(B=-1\).
Hence \[\frac{2}{r^2+4r+3}=\frac{1}{r+1}-\frac{1}{r+3}.\]
An alternative approach, starting from equation \(\eqref{eq:partial}\), is to multiply both sides by \((r+1)(r+3)\), giving \[\begin{equation}\label{eq:identity} 2=A(r+3)+B(r+1). \end{equation}\]We could proceed as before, expanding the right hand side to get the identity \(2=(A+B)r+3A+B\). Alternatively, as this identity \(\eqref{eq:identity}\) is true for all values of \(r\), we can substitute in \(r=-1\) to obtain \(2=2A\), and hence \(A=1\), and then substitute in \(r=-3\) to obtain \(2=-2B\), giving \(B=-1\).
Hence, or otherwise, prove that \[\sum_{r=1}^n \frac{2}{r^2+4r+3}=\frac{5}{6}-\frac{2n+5}{(n+2)(n+3)}.\]
For this part of the question, the idea is to use our partial fractions result to split the sum of interest into two simpler sums.
Substituting in the result from the previous part of the question, we obtain \[\begin{align*} \sum_{r=1}^n \frac{2}{r^2+4r+3} &=\sum_{r=1}^n\left(\frac{1}{r+1}-\frac{1}{r+3}\right)\\ &=\sum_{r=1}^n\frac{1}{r+1}-\sum_{r=1}^n\frac{1}{r+3}, \end{align*}\]where we have split the single sum into two separate sums on the second line.
This still looks rather intimidating, so let’s do away with the summation notation and write out the sums explicitly: \[\begin{align*} \sum_{r=1}^n \frac{2}{r^2+4r+3}&{}=\left(\frac{1}{2}+\frac{1}{3}+\boldsymbol{\frac{1}{4}+\dotsb+\frac{1}{n+1}}\right) \\ &{}-\left(\boldsymbol{\frac{1}{4}+\frac{1}{5}+\dotsb+\frac{1}{n+1}}+\frac{1}{n+2}+\frac{1}{n+3}\right). \end{align*}\] Now it is clear that the bold terms in the first sum are cancelled by corresponding terms in the second. Therefore, we are left with \[\begin{align*} \sum_{r=1}^n \frac{2}{r^2+4r+3}&{}=\left(\frac{1}{2}+\frac{1}{3}\right)- \left(\frac{1}{n+2}+\frac{1}{n+3}\right)\\ &{}=\frac{5}{6}-\frac{2n+5}{(n+2)(n+3)}. \end{align*}\]
