korth0221 said:
4. Find interval of convergence of power series n=1 to infinity: (-1^(n+1)*(x-4)^n)/(n*9^n)
$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n + 1}\left( x - 4 \right) ^n}{n\,9^n} } \end{align*}$
We can say for sure that the series is absolutely convergent where $\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\left| a_{n+1} \right| }{\left| a_n \right| } } < 1 \end{align*}$, so
$\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\left| \frac{\left( -1 \right) ^{n + 2} \left( x - 4 \right) ^{n + 1}}{\left( n + 1 \right) \, 9^{n+1}} \right| }{ \left| \frac{\left( -1 \right) ^{n + 1}\left( x - 4 \right) ^n}{n\,9^n} \right| } } &< 1 \\ \lim_{n \to \infty}{ \frac{\frac{\left| x - 4 \right| ^{n + 1}}{\left( n + 1 \right) \, 9^{n + 1}}}{\frac{\left| x - 4 \right| ^n}{n\,9^n}} } &< 1 \\ \lim_{n \to \infty}{ \frac{n\,9^n\,\left| x - 4 \right| ^{n+1}}{\left( n + 1 \right) \, 9^{n+1}\,\left| x - 4 \right| ^n} } &< 1 \\ \lim_{n \to \infty}{ \left( \frac{n}{n + 1} \right)\,\frac{\left| x - 4 \right|}{9} } &< 1 \\ \lim_{n \to \infty}{ \left( 1 - \frac{1}{n + 1} \right) \, \frac{ \left| x - 4 \right| }{9} } &< 1 \\ \left( 1 - 0 \right) \,\frac{\left| x - 4 \right| }{9} &< 1 \\ \frac{\left| x - 4 \right| }{9} &< 1 \\ \left| x - 4 \right| &< 9 \\ -9 < x - 4 &< 9 \\ -5 < x &< 13 \end{align*}$
We know for certain that the series is absolutely convergent where $\displaystyle \begin{align*} -5 < x < 13 \end{align*}$, but the ratio test tells us nothing about the convergence at the endpoints. So we will check what happens there... $\displaystyle \begin{align*} x = -5 \end{align*}$ gives
$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n+1} \left( -5 - 4 \right) ^n }{ n\,9^n } } &= \sum_{n = 1}^{\infty}{ \frac{\left( - 1 \right) ^{n + 1}\left( -9 \right) ^n }{ n\,9^n } } \\ &= \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n + 1} \left( -1 \right) ^n \, 9^n}{n\,9^n} } \\ &= \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n + 2}}{n} } \\ &= \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^n}{n} } \end{align*}$
The Alternating Harmonic Series is well known to be convergent, as $\displaystyle \begin{align*} \frac{1}{n} \end{align*}$ is decreasing. So the series converges at $\displaystyle \begin{align*} x = -5 \end{align*}$.
When $\displaystyle \begin{align*} x = 13 \end{align*}$ we have
$\displaystyle \begin{align*} \sum_{n =1}^{\infty}{ \frac{\left( -1 \right)^{n + 1}\left( 13 - 4 \right) ^n}{n\,9^n} } &= \sum_{ n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n +1} \,9^n}{n\,9^n} } \\ &= -\sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^n}{n} } \end{align*}$
again, this has been reduced to a multiple of the Alternating Harmonic Series, so we know the series is convergent at $\displaystyle \begin{align*} x = 13 \end{align*}$.
Thus the interval of convergence is $\displaystyle \begin{align*} -5 \leq x \leq 13 \end{align*}$.