I'm trying to solve this equation:
$$\sum_{k = 0}^{\infty}\dfrac{1}{(k+1)(k+3)}$$
Original image at https://i.stack.imgur.com/2WINT.png
I attempted to find the sums of
$\sum_0^∞\frac{1}{k+1}$ and $\sum_0^∞\frac{1}{k+3}$ and then attempt to multiply, but then I realized neither converges, due to the denominator power being equal to 1.
How do I do this?
On
Further hint: What is $\frac{1}{k+1} - \frac{1}{k+3}$ with a common denominator?
How does that result relate to $\frac{1}{(k+1)(k+3)}$?
What happens to each term because you're subtracting?
On
Hint: Have you seen this example?
$$\begin{align} \sum_{k=1}^\infty{1\over k(k+1)}&=\sum_{k=1}^\infty\left({1\over k}-{1\over k+1} \right)\\ &=\left({1\over1}-{1\over2}\right)+\left({1\over2}-{1\over3}\right)+\left({1\over3}-{1\over4}\right)+\cdots\\ &=1-\left({1\over2}-{1\over2}\right)-\left({1\over3}-{1\over3}\right)-\cdots \end{align}$$
Your series is similar (albeit a bit more complicated).
On
Applying partial fractions, we find $$\frac{1}{(k+1)(k+3)}=\frac{\frac{1}{2}}{k+1}-\frac{\frac{1}{2}}{k+3}$$
Hence we have
\begin{align}\sum\limits_{k=0}^\infty\frac{1}{(k+1)(k+3)}&=\frac{1}{2}\sum\limits_{k=0}^\infty\left(\frac{1}{k+1}-\frac{1}{k+3}\right) \\&=\frac{1}{2}\left[\frac{1}{1}+\frac{1}{2}+\sum\limits_{k=0}^\infty\left(\frac{1}{k+3}-\frac{1}{k+3}\right)\right] \\&=\frac{1}{2}\left[\frac{1}{1}+\frac{1}{2}\right] \\&=\frac{3}{4}\end{align}
Where in the antepenultimate step we have just changed the index of the first part of the summation.
Hint: partial fractions, telescoping series...