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.

 

[Gib Z]

Is this for our general interest or something?

 

[Ulagatin]

Oh, sorry, wrote this out for someone, ended up posting it too, and I didn't explain its purpose.

Could be taken as being for general interest if you like.