Telescoping Lawrence's Sum

$\frac {1}{n!} - \frac {1}{\left(n+1\right)!} = \frac {n+1}{\left(n+1\right)!} - \frac {1}{\left(n+1\right)!} = \frac {n+1-1}{\left(n+1\right)!} = \frac {n}{\left(n+1\right)!}$ .

I'm not sure how to tag, so sorry, Finn Hulse.

Taking 1/n! common from both we get (1/n!) (1-1/(n+1))=1/n! ((n+1-1)/(n+1))=1/n!*(n/n+1)=n/(n+1)!

I'd just use experimentation. In general, factorials are tricky to work with. Anyways, for this particular problem, I used $n=4$ . Plugging this into the equation gives $1/24-1/120 \Longrightarrow 1/30$ . Plugging $n=4$ into each one of the answer choices finds that $n/(n+1)!$ is the correct solution. I would be very interested to see a clever approach to the problem, so somebody please tag me in that solution. :D

$\frac{1}{n!} - \frac{1}{(n+1)!}$

Rewrite $(n+1)!$ as $n!(n+1$ .

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

Multiply the left term by $\frac{n+1}{n+1}$ to find a common denominator.

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

Subtract the two fractions as normal.

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

Simplify.

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

Rewrite $n!(n+1)$ as $(n+1)!$ .

$= \boxed{\frac{n}{(n+1)!}}$