Simplifying factorial in the denominator

Question

$$ \left( \frac{1}{5!} + \frac{1}{3!}\frac{1}{2!} + \frac{1}{4!} \right)z^5 - \left( \frac{1}{7!} + \frac{1}{5!} \frac{1}{2!} + \frac{1}{3!}\frac{1}{4!} + \frac{1}{6!} \right) z^7 $$

Its confusing to me how one can simplify these terms without actually calculating the factorials. Are there any laws that would help me simplify this without a calculator?

Answer 1

Hint:

$5!$ is the least common denominator for the first coefficient, so $$\dfrac1{5!}+\frac1{2!\,3!}+\frac 1{4!}=\dfrac{1+10+5}{5!}=\dotsm$$ Similarly for the second coefficient: $$5!\,2!\cdot3\cdot7=7!, \qquad 4!\cdot5\cdot7=7!$$ so $$\dfrac1{7!}+\dfrac1{6!}+\frac1{5!\,2!}+\frac 1{4!\,3!}=\frac{1+7+21+35}{7!}.$$ One may conjecture the following formula: $$\sum_{k=0}^n\frac1{(2n-k+1)k!}=\frac{2^{2n}}{(2n+1)!}.$$

Answer 2

In the first term, there is a common factor of $1/(2 \times 3!)$ since $5! = 5 \times 4 \times 3!$ and $4!= 4 \times 3!$. Factorising, $$ \frac{1}{5!} + \frac{1}{3!2!} + \frac{1}{4!} = \frac{1}{2 \times 3!} \left( \frac{1}{5} + 1 + \frac{1}{2} \right), $$ which looks okay to calculate.

The second term is worse, but there's a common factor of at least $4!$, so $$ \frac{1}{7!}+\frac{1}{6!} + \frac{1}{5!2!} + \frac{1}{4!3!} = \frac{1}{4!} \left( \frac{1}{7 \times 6 \times 5} + \frac{1}{6 \times 5} + \frac{1}{5 \times 2} + \frac{1}{3 \times 2} \right). $$ Now we can see there's another factor of $2$ that can be taken out, and you can probably carry on from here.

Answer 3

$$\bigg(\frac{1}{5!}+\frac{1}{3!}\cdot \frac{1}{2!}+\frac{1}{4!}\bigg)= \frac{2!+5\cdot 4+5\cdot 2!}{5!2!} =\frac{2(16)}{5!2}=\frac{2^4}{5\cdot 2^3\cdot 3}=\frac{2}{15}$$

$$\bigg(\frac{1}{7!}+\frac{1}{5!}\cdot\frac{1}{2!}+\frac{1}{3!}\cdot\frac{1}{4!}+\frac{1}{6!}\bigg)=\frac{3!+6\cdot 7\cdot 3+7\cdot 6\cdot 5+7\cdot 3!}{7!\cdot3!} $$

$$= \frac{3(2+6\cdot 7+7\cdot 2\cdot 5+7\cdot 2)}{7!3!} $$

$$=\frac{3!(1+3\cdot 7+7\cdot 5+7)}{7!3!} =\frac{2\cdot 2\cdot 4^2}{7!} = \frac{2\cdot 4}{7\cdot 6\cdot 5\cdot3} = \frac{4}{35\cdot 3^2} =\frac{4}{315}$$

So then you're left with $$\frac{2}{15}z^5-\frac{4}{315}z^7$$

Answer 4

Observe \begin{align} \left( \frac{1}{5!} + \frac{1}{3!}\frac{1}{2!} + \frac{1}{4!} \right)=&\ \frac{1}{5!}\left(\frac{5!}{5!0!} + \frac{5!}{3!2!} + \frac{5!}{4!1!} \right)\\ =&\ \frac{1}{2(5!)}\left(\frac{5!}{5!0!} +\frac{5!}{4!1!}+ \frac{5!}{3!2!} + \frac{5!}{2!3!}+\frac{5!}{1!4!}+\frac{5!}{0!5!}\right)\\ =&\ \frac{1}{2(5!)}\sum^5_{k=0}\binom{5}{k} = \frac{1}{2(5!)}2^5 = \frac{2^4}{5!} \end{align} and likewise \begin{align} \left(\frac{1}{7!}+\frac{1}{6!1!}+\frac{1}{5!2!}+\frac{1}{4!3!}\right) = \frac{2^6}{7!}. \end{align}

Edit: The above method proves the identity suggested by Bernard.