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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:

[tex]\sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \sum_{k=1}^n \frac{k!(k+1)}{k!(k+1)(k+2)(k+3)} [/tex]

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

[tex]\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)}[/tex]

Now consider the kth term:

[tex]U_{k} = \frac{1}{(k+2)(k+3)}[/tex]

Apply a differences method:

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

[tex]U_{k} = \frac{(k+3)}{(k+2)(k+3)} - \frac{(k+2)}{(k+2)(k+3)}[/tex]

[tex]U_{k} = \frac{1}{k+2} - \frac{1}{k+3}[/tex]

[tex].: U_{k} = V_{k} - V_{k+1}[/tex]

Note that [tex]\inline{V_{k} = \frac{1}{k+2} \forall{k} \in N}[/tex].

[tex]S_{n} = U_{1} + U_{2} + U_{3} + U_{4} + ... + U_{n}[/tex]

[tex]S_{n} = (V_{1} - V_{2}) + (V_{2} - V_{3}) + ... + (V_{n} - V_{n+1})[/tex]

[tex].: S_{n} = V_{1} - V_{n+1}[/tex]

[tex]\sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = S_{n}[/tex]

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

[tex]V_{1} - V_{n+1} = \frac{1}{3} - \frac{1}{n+3}[/tex]

[tex].: \sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \frac{1}{3} - \frac{1}{n+3}[/tex]

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

And for the sum to infinity, follow these steps:

[tex]\\lim_{n\Rightarrow {\infty}} \sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \\lim_{n\Rightarrow {\infty}} (\frac{1}{3} - \frac{1}{n+3})[/tex]

[tex]\\lim_{n\Rightarrow {\infty}} (\frac{1}{3} - \frac{1}{n+3}) = \frac{1}{3}[/tex]

[tex].: \\lim_{n\Rightarrow {\infty}} S_{n} = \frac{1}{3}[/tex]

The problem has thus been completed.