I doing some derivation for my work. In some textbook, I got the following simplification, $x = \sum_{k=0}^{L-1}\frac{1}{2p+2(k+1)}$ to $x=\frac{2^{L-1}\sum_{k=0}^{L-1}\frac{L!}{k+1}}{\prod_{k=0}^{L-1}2p+2(k+1)}$, where $p$ is constant.
I could get the denominator, but I'm not able to get how the numerator got modified? Is this true simplification using the method of mathematical induction or if its true how it can done?
Please clarify my doubt? Thanks.
is this: $$x=\frac{ \sum_{k=0}^{L-1}{\left(\prod_{i\ne k}^{L-1} {2p + 2(i+1)} \right)}}{ \prod_{j=0}^{L-1} {2p + 2(j+1)}}$$ useful?
For me seems very unlikely that the simplification holds true, there's no dependence on $p$ at the numerator where clearly should be being the denominator just the product of all the denominators without any simplification. Let us take it easy and try to prove it with a small amount of factors, like $L=3$. In this case we get $$x = \sum_{k=0}^2 \frac{1}{2p+2(k+1)} = \frac{1}{2p+2}+\frac{1}{2p+4}+\frac{1}{2p+6} \\ = \frac{(2p+4)(2p+6)+(2p+2)(2p+6)+(2p+2)(2p+4)}{(2p+2)(2p+4)(2p+6)} \\= \frac{3(4p^2)+2(2p\cdot6)+2(2p\cdot4)+2(2p\cdot2)+(4\cdot6)+(2\cdot6)+(2\cdot4)}{\prod_{k=0}^2 2p+2(k+1)} \\=\frac{3(4p^2)+p(8+16+24)+(8+12+24)}{\prod_{k=0}^2 2p+2(k+1)}$$ From your formula, for $L=3$ we have $$2^2\sum_{k=0}^2\frac{3!}{k-1} = 4\left(3!+\frac{3!}{2}+\frac{3!}{3}\right)$$ which clearly is not the same thing we got before
