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

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

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    $\begingroup$ Just bring everything to the same common denominator (which should be $n(n+1)(n+1)!$ here) $\endgroup$ – Luke Sep 14 '18 at 11:51
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Given $$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$$ $$=\frac{1}{(n+1)n!} + \frac{1}{(n+1)(n+1)n!} - \frac{1}{n(n!)}$$ $$=\dfrac{1}{n!}\left(\frac{1}{(n+1)} + \frac{1}{(n+1)(n+1)} - \frac{1}{n}\right)$$ $$=\dfrac{1}{n!}\left(-\dfrac{1}{n(n+1)}+ \frac{1}{(n+1)(n+1)}\right)$$ $$=\dfrac{-1}{n(n+1)(n+1)!}$$

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    $\begingroup$ I was stuck on the second, thank you $\endgroup$ – KEVIN DLL Sep 14 '18 at 11:57
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Hint: Use that $$(n+1)!=n!(n+1)$$

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We have that

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

$$=\left( \color{red}{ 1+ \frac{1}{n+1} - \frac{n+1}{n} }\right)\cdot\frac{1}{(n+1)!}$$

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