So here is the series
$\displaystyle \Sigma\limits_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!}$
For what values of x will it converge?
According to the ratio test, a series is convergent when $\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}$.
So get an expression for this limit, set it less than 1, then solve for x.
Right I was thinking along the lines of ratio test
I get this
$\displaystyle \lim_{n\rightarrow\infty} \frac{x^2}{2n(2n+1)}$
which becomes
$\displaystyle x^2\lim_{n\rightarrow\infty} \frac{1}{2n(2n+1)}$
clearly, the limit goes to 0 which is less than 1 so it converges. What can i say about x?
For this series,
$\displaystyle \Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}$
Clearly by ratio test, I get that
$\displaystyle \lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1$. So it's inconclusive here. What can I say about which values of x will it converge?
Maybe you should try the ratio test then...
$\displaystyle \displaystyle \begin{align*} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{\frac{(n+1)(x - 1)^{n+1}}{2^{n+1}}}{\frac{n(x-1)^n}{2^n}} \right| \\ &= \left| \frac{2^n(n+1)(x-1)^{n+1}}{2^{n+1}n(x-1)^n} \right| \\ &= \frac{1}{2}|x - 1| \left| \frac{n + 1}{n} \right| \end{align*}$
The "n" stuff goes to 1, but the rest doesn't. You should be able to get an interval for x when you set this entire limit less than 1.
Not true at all. For any series, if you can show the series of absolute values is convergent, then so is the original series. What we are doing by taking the absolute value of the ratio in the ratio test is to check the convergence of the absolute value series. When this is convergent, so is the original series.