Velvet Star Monitor

Standout celebrity highlights with iconic style.

updates

Evaluating an infinite series using partial fractions...

Writer Mia Lopez
$\begingroup$

I am having trouble evaluating an infinite series that uses partial fractions. The problem is as follows: $$ \sum_{n = 1}^{\infty}{1 \over n\left(n + 1\right)\left(n + 2\right)} $$

I realize that this is a telescoping series, but I am unable to find a general formula for the Sn.

After the partial fraction decomposition the problem looks like $1/\left(2n\right) - 1/\left(n + 1\right) + 1/\left(2n + 4\right)$. I input values for $n = 1, n = 2,$ etc, and some cancel out but I am unable to determine a pattern to write to find Sn... Thanks.

I found this answer to the problem online which makes sense, up to the step where it makes Sn equal to $1/2(1/2 - 1/\left(n+1\right) + 1/\left(n+2\right)$. I get the prior steps but I am not sure how they make that leap.

enter image description here

$\endgroup$ 2

1 Answer

$\begingroup$

$$\frac{1}{n(n+1)(n+2)} = \frac{1}{(n+1)} \cdot \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right) = \frac{1}{2} \left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$$

So

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left(\frac{1}{1(1+1)}-\frac{1}{(1+1)(1+2)}+\frac{1}{2(2+1)}-\frac{1}{(2+1)(2+2)}+\cdots\right)$$

$$=\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$$

$\endgroup$ 7

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy