# An example of a series involving factorials

Ulagatin
The problem at hand: $$\inline{\sum_{k=1}^n \frac{(k+1)!}{(k+3)!}}$$
Hence, find the limiting sum of the series, as n ---> infinity.

Start this summation by expanding out the factorial to have a common factor of k!(k+1) as follows:

$$\sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \sum_{k=1}^n \frac{k!(k+1)}{k!(k+1)(k+2)(k+3)}$$

Next step is to cancel the terms on the numerator and denominator:

$$\sum_{k=1}^n \frac{k!(k+1)}{k!(k+1)(k+2)(k+3)} = \sum_{k=1}^n \frac{1}{(k+2)(k+3)}$$

Now consider the kth term:

$$U_{k} = \frac{1}{(k+2)(k+3)}$$

Apply a differences method:

$$U_{k} = \frac{1}{(k+2)(k+3)}*[\frac{(k+3) - (k+2)}{1}]$$

$$U_{k} = \frac{(k+3)}{(k+2)(k+3)} - \frac{(k+2)}{(k+2)(k+3)}$$

$$U_{k} = \frac{1}{k+2} - \frac{1}{k+3}$$

$$.: U_{k} = V_{k} - V_{k+1}$$

Note that $$\inline{V_{k} = \frac{1}{k+2} \forall{k} \in N}$$.

$$S_{n} = U_{1} + U_{2} + U_{3} + U_{4} + ... + U_{n}$$

$$S_{n} = (V_{1} - V_{2}) + (V_{2} - V_{3}) + ... + (V_{n} - V_{n+1})$$

$$.: S_{n} = V_{1} - V_{n+1}$$

$$\sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = S_{n}$$

Now, substitute in the values k = 1 and k = (n + 1) into $$\inline{V_{k}}$$ to get the difference $$\inline{V_{1} - V_{n+1}}$$.

$$V_{1} - V_{n+1} = \frac{1}{3} - \frac{1}{n+3}$$

$$.: \sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \frac{1}{3} - \frac{1}{n+3}$$

It has therefore been shown that the answer to this sum to n terms is $$\inline{\frac{1}{3} - \frac{1}{n+3} \forall{n} \in N}$$, the result which can be proven by mathematical induction.

And for the sum to infinity, follow these steps:

$$\\lim_{n\Rightarrow {\infty}} \sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \\lim_{n\Rightarrow {\infty}} (\frac{1}{3} - \frac{1}{n+3})$$

$$\\lim_{n\Rightarrow {\infty}} (\frac{1}{3} - \frac{1}{n+3}) = \frac{1}{3}$$

$$.: \\lim_{n\Rightarrow {\infty}} S_{n} = \frac{1}{3}$$

The problem has thus been completed.