Question

Find the sum of the convergence series.


\[\sum_{k=1}^{\infty} \frac{1}{(ak+1)(ak+a+1)}\]

Where "a" is a positive integer.

Answer 1

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Answer 2

First rewrite the summation as:\[\sum_{k=1}^{\infty}\frac{1}{(ak+1)(a(k+1)+1)}.\]Next we use partial fractions to break the fraction into two. Let:\[\frac{1}{(ak+1)(a(k+1)+1)}=\frac{A}{ak+1}+\frac{B}{a(k+1)+1}\]Then:\[1=A(a(k+1)+1)+B(ak+1)\]\[\Longrightarrow 1=(aA+aB)k+(aA+A+B)\]We end up with two equations:\[aA+aB=0\]\[(a+1)A+B=1\]Solving this system yields:\[A=\frac{1}{a},B=-\frac{1}{a}\]Therefore we can rewrite the summation by plugging in these values for A and B:\[\sum_{k=1}^{\infty}\frac{1}{a}\cdot \frac{1}{ak+1}-\frac{1}{a}\cdot \frac{1}{a(k+1)+1}=\frac{1}{a}\sum_{k=1}^\infty\frac{1}{ak+1}-\frac{1}{a(k+1)+1}.\]This is a telescoping series. If:\[S_n=\sum_{k=1}^n\frac{1}{ak+1}-\frac{1}{a(k+1)+1}\]is the partial sum, then:\[S_n=\left(\frac{1}{a+1}-\frac{1}{2a+1}\right)+\left(\frac{1}{2a+1}-\frac{1}{3a+1}\right)+\cdots +\left(\frac{1}{na+1}-\frac{1}{(n+1)a+1}\right)\]\[=\frac{1}{a+1}-\frac{1}{(n+1)a+1}.\]We want the limit as the partial sum goes to infinity. This gives us:\[\frac{1}{a}\sum_{k=1}^\infty \frac{1}{ak+1}-\frac{1}{a(k+1)+1}=\frac{1}{a}\lim_{k\rightarrow\infty}S_k\]\[=\frac{1}{a}\lim_{k\rightarrow\infty}\left(\frac{1}{a+1}-\frac{1}{(k+1)a+1}\right)\]\[=\frac{1}{a}\cdot \frac{1}{a+1}=\frac{1}{a(a+1)}\]