Find the limit of the series $\sum \frac{1}{n(n+1)(n+2)}$

Question

Find the limit of the series $\sum \frac{1}{n(n+1)(n+2)}$

I tried to wite it as a combination of partial fractions.but ot didn't telescope

Answer 1

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$$

We can use factorials to rewrite the series:

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

Now we can use the Gamma Function

$$=\sum_{n=1}^{\infty} \frac{\Gamma(n)}{\Gamma(n+3)}$$

$$=\sum_{n=1}^{\infty} \frac{\Gamma(n)}{\Gamma(n+3)} \frac{\Gamma (3)}{\Gamma(3)}$$

$$=\frac{1}{\Gamma(3)} \sum_{n=1}^{\infty} \frac{\Gamma(n)\Gamma(3)}{\Gamma(n+3)}$$

Now apply the Gamma-Beta function relationship .

$$=\frac{1}{\Gamma(3)} \sum_{n=1}^{\infty} B(n,3)$$

$$=\frac{1}{\Gamma(3)} \sum_{n=1}^{\infty} \int_{0}^{1} x^{n-1}(1-x)^{3-1} dx$$

$$=\frac{1}{\Gamma(3)} \int_{0}^{1} \sum_{n=1}^{\infty} x^{n-1}(1-x)^{3-1} dx$$

$$=\frac{1}{\Gamma(3)} \int_{0}^{1} (1-x)^2 \sum_{n=1}^{\infty} x^{n-1} dx$$

We have a geometric series, and we use this to simplify:

$$=\frac{1}{\Gamma(3)} \int_{0}^{1} (1-x) dx$$

$$=-\frac{1}{2!} \left(\frac{(1-1)^2}{2}-\frac{(1-0)^2}{2} \right)$$

$$=\left(\frac{1}{2} \right) \left(\frac{1}{2}\right)$$

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