In my Calculus 2 course, we were asked to make the expression:

an=1(n+k)!

a_n = \frac{1}{(n + k)!}

into a telescoping sequence. In the past we have done things like:

12n=12n−1−12n

\frac{1}{2^n} = \frac{1}{2^{n-1}} – \frac{1}{2^n}

What I am trying to figure out is how to get the uppermost expression to look like the one just above. It is an infinite sum of an with n being the variable and k being some fixed natural number. Any input would be greatly appreciated.

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2 Answers

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Something like this?

1n!=n+1(n+1)!=n(n+1)!+1(n+1)!1(n+1)!=1n!−n(n+1)!1(n+1)!=1n!−(n+1)(n−1)!\frac {1}{n!} = \frac {n+1}{(n+1)!} = \frac {n}{(n+1)!} + \frac {1}{(n+1)!}\\

\frac {1}{(n+1)!} = \frac {1}{n!} – \frac {n}{(n+1)!}\\

\frac {1}{(n+1)!} = \frac {1}{n!} – \frac {(n+1)}{(n-1)!}\\

How would k be implemented?

– E. Wright

Oct 20 at 20:48

1(n+k)!=∞∑j=n+k(1j!−1(j+1)!)=∞∑j=n+kj(j+1)!.\begin{align}

\frac1{(n+k)!} &= \sum_{j=n+k}^\infty\left( \frac1{j!}-\frac1{(j+1)!}\right)

\\&=\sum_{j=n+k}^\infty\frac{j}{(j+1)!}.

\end{align}