# Simplifying $\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$ [closed]

How can I prove this equality: $$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}= \frac{-1}{n(n+1)(n+1)!}$$

## closed as off-topic by Namaste, Gibbs, José Carlos Santos, Adrian Keister, HoloSep 14 '18 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Gibbs, José Carlos Santos, Adrian Keister, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.

• Just bring everything to the same common denominator (which should be $n(n+1)(n+1)!$ here) – Luke Sep 14 '18 at 11:51

Given $$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$$ $$=\frac{1}{(n+1)n!} + \frac{1}{(n+1)(n+1)n!} - \frac{1}{n(n!)}$$ $$=\dfrac{1}{n!}\left(\frac{1}{(n+1)} + \frac{1}{(n+1)(n+1)} - \frac{1}{n}\right)$$ $$=\dfrac{1}{n!}\left(-\dfrac{1}{n(n+1)}+ \frac{1}{(n+1)(n+1)}\right)$$ $$=\dfrac{-1}{n(n+1)(n+1)!}$$
Hint: Use that $$(n+1)!=n!(n+1)$$
$$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}=\color{red}1\cdot \frac{1}{(n+1)!} + \color{red}{\frac{1}{n+1}}\cdot \frac{1}{(n+1)!} - \color{red}{\frac{n+1}{n}}\cdot\frac{1}{(n+1)!}=$$
$$=\left( \color{red}{ 1+ \frac{1}{n+1} - \frac{n+1}{n} }\right)\cdot\frac{1}{(n+1)!}$$