Finding Convergeance sum for two power-series.

Question

I'm starting a class on Advanced Mathematics I next semester and I found a sheet of the class'es 2012 final exams, so I'm slowly trying to solve the exercises in it or find the general layout. I will be posting a lot of questions with the exercises I find challenging, and I would like to ask for any help or methodologies that will make it easier for me to solve.

I understand the whole ordeal is categorized as "homework", but any assistance would be greatly appreciated, as I'm completely clueless and I would like to be prepared. With your help I've managed to do 4 exercises so far :-)

The following exercise is $Ex. 7$, graded for $5\%$.

Find the convergence Sum of the following two power series:

$\alpha$) $\sum\limits_{n=1}^{\infty}\dfrac{3^n}{n^2}x^n$ $\;\;\;\;\;\;\;\;$ $\beta$) $\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{3}4^n}(x-2)^n$

Again, are there any methodologies that set me on solving these? I've studied the definition of power series and convergeance (If I'm translating this correctly) through the class'es webpage, but there were no exercises in which they showcased solving.

Answer 1

Note that in both sums, you can deduce the problem - in the first one write $y=3x$ and get that is is sufficient to find the sum of $\dfrac{y^n}{n^2}$. You can make a similar argument in the second one.

To find this kind of sums, you need to use power-series and their properties, such as their integration and differentiation. Note that:

$\sum\limits_{n=1}^{\infty}\dfrac{y^n}{n^2} = \sum\limits_{n=1}^{\infty}\dfrac{1}{n}\dfrac{y^n}{n} = \sum\limits_{n=1}^{\infty}\int(\dfrac{1}{n}y^{n-1})dy = \int(\sum\limits_{n=1}^{\infty}\dfrac{1}{n}y^{n-1})dy = \int(\dfrac{1}{y}\sum\limits_{n=1}^{\infty}\dfrac{y^n}{n})dy = \int\dfrac{1}{y}\sum\limits_{n=1}^{\infty}\int y^{n-1}dy)dy = \int (\dfrac{1}{y}\int ( \sum\limits_{n=1}^{\infty}y^{n-1})dy)dy = \int (\dfrac{1}{y}\int ( \dfrac{1}{1-y})dy)dy = \dots$

This is true according to integration in power-series, which can be excluded from the sum and vice versa. From here it is just integrating simple functions.

*this is of course true for |y|<1.