The problem involves demonstrating the divergence of the series $$\frac{(-1)^nn!}{z^n}$$ using the ratio test. The context is within the study of series convergence and divergence, particularly in relation to complex numbers.
The discussion is ongoing, with participants providing guidance on how to approach the limit and questioning the assumptions made about the terms involved. There is acknowledgment of a correction regarding the absolute value in the denominator, and some participants are exploring the implications of the magnitude of complex numbers.
There is a mention of the importance of the absolute value in the denominator and the implications of the growth of factorial terms in the numerator. The discussion reflects on the behavior of the series as \( n \) approaches infinity, with varying interpretations of the limit's behavior.
Hi vbrasic:vbrasic said:I'm having trouble showing why the limit of this must be greater than one.
vbrasic said: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.
vbrasic said: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?