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$a_n=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n!}{2^n}$, I am trying to prove it using Leibniz theorem, Condition (i) - all $u_n$ are positive is satisfied, Condition (ii) $u_n\geq u_{n+1}$, here how do we prove that $\frac{n!}{2^n}\geq\frac{(n+1)!}{2^{n+1}}$. To prove that a function is decreasing we take derivative and prove it that for some $x=x_0$ to $x\to\infty$ the function deceases and we have condition (ii) satisfied. Taking derivative is simple but I think is not possible always as in this case where I don't know how to find the slope of $f(x)=\frac{x!}{2^n}$ with basic calculus, whereas knowing how to prove $u_n\geq u_{n+1}$ will help a lot.

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$\frac{(n+1)!}{2^{n+1}}=\frac{n!}{2^n}\cdot\frac{n+1}2>\frac{n!}{2^n}$ whenever $n>1$, so you certainly can’t prove that $\frac{(n+1)!}{2^{n+1}}\le\frac{n!}{2^n}$ in general. – Brian M. Scott Apr 8 '12 at 5:08
The general term is not decreasing! The series is non convergent. Maybe you meant the reciprocal. – Pedro Tamaroff Apr 8 '12 at 22:23

You can"t prove what you want. You can use D'Alembert (the absolute value version) .

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And in this way you will see that the series is divergent. – alpha.Debi Apr 9 '12 at 7:47

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