Convergence of Ratio Test for Complex Numbers z in C

[Dustinsfl]

$$\sum\limits_{n = 0}^{\infty}\frac{2\pi^{n+1}M}{(n+1)!}z^n$$

So we get $\lim\limits_{n\to\infty}\left|\frac{\pi z^n}{n+2}\right|$.

This converges but I don't see how. z is in C.

 

[Evgeny.Makarov]

Ratio test involves $\left|\frac{a_{n+1}}{a_n}\right|$ where $a_n$, $a_{n+1}$ are coefficients of the series. In other words, $z$ should not be in the ratio.

 

[Dustinsfl]

Ratio test involves $\left|\frac{a_{n+1}}{a_n}\right|$ where $a_n$, $a_{n+1}$ are coefficients of the series. In other words, $z$ should not be in the ratio.

Even though z has a power of n?

---------- Post added at 08:07 PM ---------- Previous post was at 07:21 PM ----------

Even though z has a power of n?

I see my mistake. It should be z not z^n

 

[Evgeny.Makarov]

It should be z not z^n

In a power series $\sum a_nz^n$, the numbers $a_n$ are called coefficients and the products $a_nz^n$ are called terms. The ratio test involves the ratio of coefficients, not terms. The ratio $a_{n+1}/a_n$ has nothing to do with $z$.

 

[chisigma]

The ratio test of a series $\displaystyle \sum_{n=0}^{\infty} a_{n}$ is a test on the quantity $\displaystyle r_{n}=|\frac{a_{n+1}}{a_{n}}|$. In this case is $\displaystyle a_{n}= \frac{2\ \pi^{n+1}\ M}{(n+1)!}\ z^{n}$ so that $\displaystyle r_{n}= |\frac{\pi\ z}{n+2}|$...

Kind regards

$\chi$ $\sigma$

 

[Evgeny.Makarov]

OK, I had a momentary lapse of reason. Of course, ratio test is a test for any series, not just power series, so the concept of coefficient does not apply here. You are right, chisigma and dwsmith.