Is $\sum_{n=1}^{\infty}(-1)^nn!2^{-n}$ a divergent series?

Question

Determine whether or not the series converge: $\sum_{n=1}^{\infty}\frac{(-1)^nn!}{2^n}$

I did do the ratio test to determine that it diverges but is there a way to not use this method as it isn't part of my course's content? Would just like a hint to how to go about this.

Thanks in advance.

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What I've tried:

I can see that $\frac{(-1)^nn!}{2^n}=(-\frac{1}{2})^nn!$ and can see a geometric series that converges but it's multiplied by an unfixed number $n!$. Can this lead to anything?

Answer 1

Let

$$f(n) = \frac{n!}{2^n} \tag{1}\label{eq1A}$$

You have

$$f(4) = \frac{4!}{2^4} = \frac{24}{16} = \frac{3}{2} \gt 1 \tag{2}\label{eq2A}$$

Since each additional term is the previous term multiplied by $\frac{n+1}{2} \gt 1$, you can prove quite easily by induction (which I'll leave you to do) that

$$f(n) \gt 1, \; \forall \; n \ge 4 \tag{3}\label{eq3A}$$

Since the individual terms don't converge to $0$, the sum can't converge either, even as an alternating series, i.e., it diverges.

Answer 2

Using Stirling's approximation

$$\sqrt[n]{\lim_{n\to\infty}\left(-\dfrac12\right)^nn!}=\sqrt[n]{\lim_{n\to\infty}\left(-\dfrac12\right)^n\sqrt{2n\pi}\left(\dfrac ne\right)^n}=\lim_{n\to\infty}\left(-\dfrac n{2e}\right)(\sqrt{2\pi n})^{1/n}=?$$

Use this $$\lim_{n\to\infty}n^{1/n}=1$$

Answer 3

$n!/2^n\geq n/4.$ Can nth term of the given series converge to zero, which is the necessary condition for the series to converge?

Answer 4

Let $n > 4$;

$n!=n(n-1)(n-2)......1$;
$n$ factors.

$n!/2^n=$

$(n/2)((n-1)/2)((n-2)/2)...........(2/2)(1/2)\gt$

$(n/4)\cdot 1\cdot 1.......\cdot 1\gt 1$.

Hence $\lim_{n \rightarrow \infty }n!/2^n >1 \not =0$,

the series diverges.