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I need help finishing this proof. I've come to a point where I don't know how to continue. I need to prove that the following inequality is true for all positive integers $n$.

$$\sum_{i=1}^{n-1} (-1)^{i+1}i! \leq \frac{(2n)!}{2}$$

It's true for $P(1)$, now I consider the inequality to be my inductive hypothesis, I want to show that $\sum_{i=1}^{n} (-1)^{i+1}i! \leq \frac{(2n+2)!}{2}$ it's true for $(n+1)$. Taking my i.h.

$$\sum_{i=1}^{n-1} (-1)^{i+1}i! + (-1)^{n+1}n! \leq \frac{(2n)!}{2}+ (-1)^{n+1}n!$$

I should consider even and odd n's differently, when I consider an odd $n$

$$\sum_{i=1}^{n-1} (-1)^{i+1}i! + n! \leq \frac{(2n)!}{2}+ n!$$

So I should check that $\frac{2n!}{2}+ n! \leq \frac{(2n+2)!}{2}$ right? This takes me to:

$$\frac{(2n)!}{2}+ \frac{2(n!)}{2} \leq \frac{(2n+2)!}{2}$$

$$(2n)!+ 2(n!) \leq (2n+2)!$$

Am I on the right path here? How can I prove this inequality? Should I make an auxiliar proof by induction or is it self-evident as it is?

Thanks.

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    $\begingroup$ Hint: $(2n)!+ 2(n!) \leq (2n)!+ 2(\color{red}{2}n)! \leq 3(2n)!$ $\endgroup$ – Steve Kass May 10 '16 at 18:14
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You’re doing fine, though you don’t need to treat even and odd $n$ separately: clearly $\frac{(2n)!}2-n!<\frac{(2n)!}2+n!$, so it’s sufficient to show that you always have

$$\frac{(2n)!}2+n!\le\frac{(2n+2)!}2$$

or, equivalently, that

$$(2n)!+2n!\le(2n+2)!\;.$$

HINT: Show that $2n!\le(2n)!$, so that $(2n)!+2n!\le 2(2n)!$, and then compare this with $(2n+2)!$.

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That inequality is quite weak. We have to bound: $$ \sum_{i=1}^{n-1} (-1)^{i+1} i! = (-1)^n \sum_{j=1}^{n-1}(-1)^{j+1}(n-j)! $$ but the terms of the last sum are decreasing in absolute value, hence $\sum_{j=1}^{n-1}(-1)^{j+1}(n-j)!$ is between $(n-1)!=(n-1)\cdot(n-2)!$ and $(n-1)!-(n-2)! = (n-2)\cdot (n-2)!$.

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