Evaluating a simple sum

Question

I have a question that $\sum _{k =1}^n\frac{1} {k(k+3)} $ I resolved this into partial fraction $\sum \left(\frac{1} {3k} - \frac{1} {3(k+3)} \right) $

I had a geometric series which converges to $1/2$ but the other has a different pattern

Where exactly is the way to go?

Answer 1

You can rewrite it further:

$$\frac{1}{3} \sum_{k=1}^{n}(\frac{1}{k} - \frac{1}{k + 3}) = \frac{1}{3}(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} - \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3}) = \frac{1}{3}(\frac{11}{6} - \frac{3n^2+12n+11}{(n+1)(n+2)(n+3)})$$

Other elements of the sum are cancelled.

In a special case

$$\lim_{n \to \infty}\sum_{k=1}^n(\frac{1}{3k} - \frac{1}{3(k+3)})=\frac{11}{18}$$

Answer 2

After rewriting with partial fractions, the resulting series telescopes. Write out the first several terms, and you’ll see the massive cancellation.

Answer 3

$$\sum _{k =1}^n\frac{1} {k(k+3)}=\\ \sum _{k =1}^n\frac{1\times(k+1)(k+2)} {k(k+1)(k+2)(k+3)}=\\\ \sum _{k =1}^n\frac{\frac{1}{3}(k+3-k)(k+1)(k+2)} {k(k+1)(k+2)(k+3)}=\\ \\ \frac{1}{3}\sum _{k =1}^n\frac{((k+3)-(k))(k+1)(k+2)} {k(k+1)(k+2)(k+3)}=\\ \frac{1}{3}\sum _{k =1}^n\frac{(k+1)(k+2)(k+3)-k(k+1)(k+2)} {k(k+1)(k+2)(k+3)}=\\ \frac{1}{3}\sum _{k =1}^n\frac{(k+1)(k+2)(k+3)-k(k+1)(k+2)} {k(k+1)(k+2)(k+3)}=\\ \frac{1}{3}\sum _{k =1}^n\frac{(k+1)(k+2)(k+3)} {k(k+1)(k+2)(k+3)}-\frac{k(k+1)(k+2)} {k(k+1)(k+2)(k+3)}=\\ \frac{1}{3}\sum _{k =1}^n \frac{1} {k}-\frac{1} {k+3}\\ \frac{1}{3}\sum _{k =1}^n (\frac{1} {k}-\frac{1} {k+1})+(\frac{1} {k+1}-\frac{1} {k+2})+(\frac{1} {k+2}-\frac{1} {k+3})\\$$now you can take over