series...again..and again

[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)}

[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}

[Alem2000]

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

[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

[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).