Determine for which values of $x$ the series $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n^2}$ converges.
Use the Ratio Test to determine what the radius of convergence should be for this series.
The radius of convergence here is $R=1$, so we know for sure that the series converges absolutely for $|x| < 1$.
Next, we have to check the two endpoints at $x=1$ and $x=-1$. For these, we get the series $\displaystyle \sum_{n=0}^\infty \frac{1}{n^2}$ and $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n^2}$. Do these converge?
Both of these series converge absolutely.
The interval of convergence here is $[-1,1]$.
Determine for which values of $x$ the series $\displaystyle \sum_{n=0}^\infty \left(\frac{4}{3}\right)^n (x-2)^n$ converges.
Use the ratio test to determine what the radius of convergence should be.
The radius of convergence is $R = \frac{3}{4}$ so this series converges when $|x-2| < \frac{3}{4}$.
Next, we need to check the endpoints. What happens if $x = 2+\frac{3}{4}$ and $x = 2 - \frac{3}{4}$?
Those series become $\displaystyle \sum_{n=0}^\infty 1^n$ and $\displaystyle \sum_{n=0}^\infty (-1)^n$, both of which diverge.
The interval of convergence here is $\left(\frac{5}{4}, \frac{7}{4}\right)$.
Determine for which values of $x$ the series $\displaystyle \sum_{n=0}^\infty \frac{1}{n2^n}(x+1)^n$ converges.
Use the ratio test to try to figure out the radius of convergence.
The radius of convergence is $2$ here, so we know the series for $|x+1| < 2$.
Next, we need to check the endpoints, which are at $x=-3$ and $x=1$. What do these series look like?
The series we need to deal with are $\displaystyle \sum_{n=0}^\infty \frac{1}{n}$ and $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n}$
The interval of convergence here is $[-3, 1)$.
Determine for which values of $x$ the series $\displaystyle \sum_{n=0}^\infty \frac{n}{5^n}x^{2n}$ converges.
Use the ratio test to try to figure out the radius of convergence. Be careful with the $x^{2n}$ when working with the ratio test.
The radius of convergence is $\sqrt{5}$ here, so we know the series for $|x| < \sqrt{5}$.
Next, we need to check the endpoints, which are at $x=-\sqrt{5}$ and $x=\sqrt{5}$. What do these series look like?
The series we need to deal with are $\displaystyle \sum_{n=0}^\infty n$ and $\displaystyle \sum_{n=0}^\infty n(-1)^n$
The interval of convergence here is $(-\sqrt{5}, \sqrt{5})$.
Find a power series expansion for the function $\frac{1}{4+3x}$ and determine the interval of convergence.
To get a power series expansion, we need to write this function in the form $\frac{1}{1- u}$ for some $u$.
The function can be rewritten as $\displaystyle \frac{1}{4} \frac{1}{1 - (-\frac{3}{4}x)}$.
Thus, the power series for this function is $\displaystyle \frac{1}{4} \sum_{n=0}^\infty \left(-\frac{3}{4}x\right)^n$
For finding the interval of convergence, we know this series converges whenever $|u|<1$, and here, we have $u = -\frac{3}{4}x$
The power series expansion is $\displaystyle \frac{1}{4} \sum_{n=0}^\infty \left(-\frac{3}{4}\right)^nx^n$, and is valid for $|x| < \frac{4}{3}$.
Find a power series expansion for the function $\displaystyle \frac{2}{5 - x}$ centered at $x = -1$ and determine the interval where it is valid.
The main trick here is that we know the power series for $\frac{1}{1-u}$ centered at 0. Since $\displaystyle \frac{1}{1-u} = \sum_{n=0}^\infty u^n$, this means that, for example, $\displaystyle \frac{1}{1 - (x+1)} = \sum_{n=0}^\infty (x+1)^n$, which is a power series centered at -1.
This means the expression we are aiming for is of the form $\displaystyle \frac{1}{1 - a(x+1)}$ for some constant $a$. This will allow the power series to be centered at $-1$.
We can rewrite $\displaystyle \frac{2}{5-x} = 2 \frac{1}{6 - (x+1)} = \frac{1}{3} \frac{1}{1 - \frac{1}{3}(x+1)}$.
Thus, the power series we get is $\displaystyle \frac{1}{3} \sum_{n=0}^\infty \left(\frac{1}{3}(x+1)\right)^n$ and it is valid where $\left|\frac{1}{3}(x+1)\right| < 1$.
The power series is $\displaystyle \frac{1}{3} \sum_{n=0}^\infty \frac{1}{3^n}(x+1)^n$ and is valid on $(-4, 2)$.
Determine for which values of $x$ the series $\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(x+2)^{n}$ converges.
Find a power series expansion for the function $\frac{2x}{(1 - 3x^2)^2}$ centered around $x=0$ and determine the interval of convergence. Hint: This is the derivative of a function with a nice power series expansion.