From $ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} $ to $ \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}} $?

Question

Good evening to everyone. I have an expression that I don't know how to arrive at. $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} $$ to $$ \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}} $$ What I've tried: $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} = \frac{n!\left(n+\frac{1}{n!}\right)\left(n+1\right)}{n!\left(\left(n+1\right)^2+\frac{1}{n!}\right)} = \frac{\left(n+\frac{1}{n!}\right)\left(n+1\right)}{\left(\left(n+1\right)^2+\frac{1}{n!}\right)} $$ And from here I don't know what to do anymore. The second attempt: $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} = \frac{\left(n^2\cdot \:\:n!+n\cdot n!+n+1\right)}{n!\left(n+1\right)^2+1} = \frac{n!\left(n^2+n+\frac{n}{n!}+\frac{1}{n!}\right)}{n!\left(\left(n+1\right)^2+\frac{1}{n!}\right)} =\frac{\left(n^2+n+\frac{n}{n!}+\frac{1}{n!}\right)}{\left(\left(n+1\right)^2+\frac{1}{n!}\right)} $$ And again I don't know what to do anymore. Thanks for any response.

Answer 1

The expression is not quite right. It is not true that $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} = \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}}. $$ For example, if you plug in $n = 1$ you get $\frac{4}{5}$ and $\frac{2}{3}$, not equal.

What is true is that $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} = \frac{n+\frac{1}{n!}}{n+1+\color{red}{\frac{1}{(n+1)!}}}. $$ Your work gets you most of the way there. You want to factor out $(n+1)!$ from the top and bottom, instead of $n!$.