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Can someone help what test to use in this equation to find the Radius of convergence and the interval of convergence?

$\displaystyle\sum (-1)^n\frac{(x+2)^n}{n}$

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    I feel Kleon needs $t$o elaborate on his/her difficulties before it is appropriate to post an answer. Just my opinion, though.2012-04-15

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To find the radius of convergence of a power series, you can use the following:

The radius of convergence of the power series $ \sum_{n=1}^\infty a_n (x-a)^n$ is $r=\lim\limits_{n\rightarrow\infty}{ |a_n|\over| a_{n+1}|}$, provided this limit exists (note the absolute values). $r=\infty$ is allowed here. (There is a Root Test version of this also.)


Recall what the radius of convergence is and what the interval of convergence is:

If the radius of convergence of the power series $\sum\limits_{n=1}^\infty a_n(x-a)^n$ is $r$, then the series converges whenever |x-a| and diverges whenever $|x-a|>r$.

The interval of convergence of a power series is the set of all $x$ for which the series converges.


To find the interval of convergence, you first find the radius of convergence and apply the following:

If $r$ is the radius of convergence of the power series $\sum\limits_{n=1}^\infty a_n(x-a)^n$, then:

$\ \ \ $1) If $r=\infty$, the interval of convergence is $(-\infty,\infty)$.

$\ \ \ $2) If $r=0$, the interval of convergence is just the singleton set $\{a\}$.

$\ \ \ $3) Otherwise, for $r\ne0$ finite, the interval of convergence is determined up to two points: the interval of convergence is the interval $(a-r,a+r)$ together with possibly one, both or neither of its endpoints.

In case 3), you have to explicitly write down the series obtained at the endpoints $x=a-r$ and $x=a+r$ and ''see what happens". That is, for example, write down the series obtained when $x=a-r$ and determine if it converges or not. If it does, then of course $a-r$ is in the interval of convergence. Do the same for the other endpoint $a+r$.



For your series, $\tag{1}\sum\limits_{n=1}^\infty (-1)^n {(x+2)^n\over n},$ you can verify, using the test above, that the radius of convergence is $r=1$.

Thus, since we have $a=-2$, the series converges whenever |x+2|<1 and diverges whenever $|x+2|>1$. So, translating this to intervals, the series converges for all $x$ in $(-3,-1)$ and diverges for all $x\notin [-3,-1]$. Thus you know the interval of convergence $I$ satisfies $ (-3,-1)\subseteq I\subseteq [-3,-1]. $ So we know what $I$ is, except that we do not know if $x=-3\in I$ or if $x=-1\in I$. We can determine whether $I$ contains either of these points or not by writing the series obtained by setting $x=-3$ or $x=-1$ in $(1)$ and seeing what happens:

For $x=-1$, the series $(1)$ becomes $ \sum_{n=1}^\infty (-1)^n {1\over n}. $ This series is a convergent alternating series. So $-1\in I$.

For $x=-3$, the series in $(1)$ becomes $ \sum_{n=1}^\infty (-1)^n {(-1)^n\over n}=\sum_{n=1}^\infty {1\over n}. $ This series is the divergent Harmonic series. So $-3\not\in I$.

Thus the interval of convergence is $I=(-3,-1]$.

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    Ah, makes sense. Thanks.2012-11-07