# Series againand again

1. Nov 9, 2004

### Alem2000

Sorry about the title everyone but ive posted numerous threads on series and I had to choose an apropriate title :tongue2:
The problem asks to use the ratio test, and determine for which values of x the test is conclusive-either converging or diverging. Then check those cases where the test is inconclusive by some other means.

here is the the series $$\sum_{n=3}^{\infty}\frac{x^n}{n3^n}$$...converge or diverge here is what i did $$\frac{a_{n+1}}{a_n}$$ and that came out to be $$\frac{x^{n+1}}{(n+1)(3^{n+1})}$$ multiplie by the $$\frac{n3^{n}}{x^{n}}$$ and after you cross out similar variables and it comes out to be

$$\lim_{x\rightarrow \infty}\frac{xn}{3(n+1)}$$

Last edited: Nov 9, 2004
2. Nov 9, 2004

### Galileo

$$\frac{a_{n+1}}{a_n}=\frac{x^{n+1}n3^n}{x^n(n+1)3^{n+1}}=\frac{nx}{(n+1)3}$$

3. Nov 9, 2004

### Alem2000

thanks galileo but I got that far just had problems latexing it

4. Nov 9, 2004

### Justin Lazear

So you get

$$\lim_{n \to \infty} \left| \frac{n}{(n+1)}\frac{x}{3}\right| = \frac{|x|}{3}$$

And you know this is stricly less than 1 for it to be conclusively converging, strictly greater than 1 to be conclusively diverging, and inconclusive at 1. So you must test the values for which the expression equals one.

--J

5. Nov 10, 2004

### James R

The remaining cases are x=3 and x=-3

If x=3 then the series reduces to the harmonic series, which diverges.

If x=-3 then we have an alternating series, and we can use the Leibnitz test (whose exact conditions escape me right now).