Show that a series is divergent

[vbrasic]

Homework Statement

Show that $$\frac{(-1)^nn!}{z^n}$$ is divergent.

Homework Equations

We can use the ratio test, which states that if, $$\lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|>1$$ a series is divergent.

The Attempt at a Solution

Applying the ratio test, we find that $$\bigg|\frac{a_{n+1}}{a_n}\bigg|=\frac{n+1}{z}.$$ I'm having trouble showing why the limit of this must be greater than one.

 

[Buzz Bloom]

I'm having trouble showing why the limit of this must be greater than one.

Hi vbrasic:

I suggest you think about the numerator and he denominator separately with respect to which gets larger than the other as n changes.

Regards,
Buzz

 

[Ray Vickson]

Homework Statement

Show that $$\frac{(-1)^nn!}{z^n}$$ is divergent.

Homework Equations

We can use the ratio test, which states that if, $$\lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|>1$$ a series is divergent.

The Attempt at a Solution

Applying the ratio test, we find that $$\bigg|\frac{a_{n+1}}{a_n}\bigg|=\frac{n+1}{z}.$$ I'm having trouble showing why the limit of this must be greater than one.

Your ration $|a_{n+1}/a_n|$ is wrong: you need $|z|$ in the denominator (and that is crucial, not just a minor quibble).

 

[vbrasic]

Added in the missing absoute value. I think the reason it must be greater than one in the limit is because for any complex number, we may write it as $re^{i\phi},$ with magnitude $r$. Given then that $r$ is finite, we have that the limit tends to $\infty$ because of the $n$ in the numerator. Does that sound okay?

 

[Ray Vickson]

Added in the missing absoute value. I think the reason it must be greater than one in the limit is because for any complex number, we may write it as $re^{i\phi},$ with magnitude $r$. Given then that $r$ is finite, we have that the limit tends to $\infty$ because of the $n$ in the numerator. Does that sound okay?

Yes, it's OK. But is would be easier to just say that for any given $z$, the quantity $|z|$ is some positive finite number, and eventually all the $n+1$ well exceed it.