$$ 0!+1!+...+n! \leq n!\bigg(\dfrac{1}{0!}+\dfrac{1}{1!}+...+\dfrac{1}{n!}\bigg) $$
How can I prove this inequality for $n\geq 1$? Actually I can see that when you divide both sides with $n!$ lhs is never greater than $2$ and rhs is always greater than or equal to $2$. I mean, $P(n) \leq 2 \leq G(n)$. But obviously that wouldn't be a proof. I am too lost in in the inductive step. How can I derive $P(n+1)$ from $P(n)$ to prove this inequality?
What I tried was adding both sides $(n+1)!$ but I couldn't make my way through that. Now I'm multiplying both sides with $(n+1)$. It's going like this
Suppose
$$ 0!+1!+...+n! \leq n!\bigg(\dfrac{1}{0!}+\dfrac{1}{1!}+...+\dfrac{1}{n!}\bigg) $$
Multiply with (n+1) to get
$$ 0!(n+1)+1!(n+1)+...+(n+1)! \leq (n+1)!\bigg(\dfrac{1}{0!}+\dfrac{1}{1!}+...+\dfrac{1}{n!}\bigg) $$ But this means
$$ 0!+1!+...+n!+(n+1)! \leq (n+1)!\bigg(\dfrac{1}{0!}+\dfrac{1}{1!}+...+\dfrac{1}{n!}\bigg) $$
As you can see I got the lhs in the form of P(n+1) but still on my way for the rhs.
LHS is $\sum_o^nk!$, RHS is $\sum_0^n\frac{n!}{k!}=\sum_0^n\frac{n!}{(n-k)!}$.
Note that $\frac{n!}{(n-k)!}=k!$$n\choose k$$\geq k!$, so the RHS is greater term-by-term.
Hence, it is also greater overall.
Induction proof:
Let $L(n)=0!+1!+...+n!, R(n)=n!(\frac{1}{0!}+\frac{1}{1!}+...\frac{1}{n!})$.
Then, $L(n+1)=L(n)+(n+1)!$, $R(n+1)=(n+1)R(n)+1$.
Then, $R(n+1)-L(n+1)=R(n)-L(n)+[nR(n)+1-(n+1)!]$
Noting that $R(n)\geq 2(n)!$ for $n\geq1$, then $nR(n)+1-(n+1)!\geq2n(n)!+1-(n+1)!=n!(2n+\frac{1}{n!}-(n+1))=n!(n-(1-\frac{1}{n!}))\geq0$
As such, $R(n)-L(n)$ grows with $n$, and as it's equal to $0$ for $n=0,1$, we see that $R(n)-L(n)\geq0$ for all $n$, whence $R(n)\geq L(n)$ and the inequality is proved.