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Prove that the sequence



My only idea was to use that $n^n \geq n!$ , but it didnt help much.

Thanks in advance.

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Have you tried comparing the size of successive terms in the sequence? – Travis Aug 25 '14 at 16:47
$$\frac{(-2)^{n!}}{n^n}=\left(-\frac{2^{(n-1)!}}n\right)^n$$ – lab bhattacharjee Aug 25 '14 at 16:47
@Giiovanna there's a \leq that should be \geq – pppqqq Aug 25 '14 at 16:48
Note that since $n!$ is almost always even, the minus sign is just a potential distraction. – alex.jordan Aug 25 '14 at 16:48

$\left|\dfrac{(-2)^{n!}}{n^n}\right| = \dfrac{2^{n!}}{n^n} = (\dfrac{2^{(n-1)!}}{n})^n$.

Remark that $\lim_{n\to +\infty}\dfrac{2^{(n-1)!}}{n} = +\infty$

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Does the divergence of the modulus of the sequence guarantee the divergence of the sequence? – Giiovanna Aug 25 '14 at 16:58
@Giiovanna yes, if a series is convergent, then its n-th term converges to 0 – Petite Etincelle Aug 25 '14 at 17:00
Serie? I didint understand, can you explain? I could prove the divergence of the modulus, but not of the sequence itself. – Giiovanna Aug 25 '14 at 17:03
@Giiovanna just a synonym for sequence – Petite Etincelle Aug 25 '14 at 17:04
@Giiovanna: If a sequence $a_n$ converges to a limit $L$ then the inequality $||a_n| - |L|| \leq |a_n - L|$ implies that $|a_n|$ converges to $|L|$, so if the modulus does not converge, then neither does the sequence. – Bungo Aug 25 '14 at 17:12

For $n \geq 3$ we have

$$\left| \frac{(-2)^{n!}}{n^n} \right| =\left(\frac{2^{(n-1)!}}{n} \right)^n $$

Now, by Bernoulli inequality $$\frac{2^{(n-1)!}}{n}=\frac{(1+1)^{(n-1)!}}{n}\geq \frac{1+1\cdot(n-1)!}{n}\geq \frac{(n-1)!}{n} \geq \frac{(n-1)(n-2)}{n}\geq n-3$$

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