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[SOLVED] Ratio test

dwsmith

Well-known member
Feb 1, 2012
1,673
$$\sum\limits_{n = 0}^{\infty}\frac{2\pi^{n+1}M}{(n+1)!}z^n$$

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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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

Well-known member
Feb 13, 2012
1,704
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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.