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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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    Ratio test? What difficulty are you having?2012-04-15
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    You can use the ratio test. It yields to it in the usual manner. Just don't forget the absolute values! (That is, remember that the ratio test begin by taking the limit of $\frac{|a_{n+1}|}{|a_n|}$, not of $\frac{a_{n+1}}{a_n}$).2012-04-15
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    I feel Kleon needs to 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 $|x-a|<r$ 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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    Why did you put $r=\lim\limits_{n\rightarrow\infty}{ |a_n|\over| a_{n+1}|}$ instead of $r=\lim\limits_{n\rightarrow\infty}{ |a_{n+1}|\over| a_{n}|}$?2012-11-07
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    @Joe By the Ratio Test, the series $\sum a_n x^n$ converges if $$\lim_{n\rightarrow\infty} {|a_{n+1} x^{n+1}|\over |a_n x^n|} = |x|\lim_{n\rightarrow\infty} {|a_{n+1}|\over |a_n |}<1.$$ So for convergence, you want $$|x|<1/ \lim_{n\rightarrow\infty} {|a_{n+1} |\over |a_n |} =\lim_{n\rightarrow\infty} {|a_n|\over|a_{n+1}|}$$ (interpreting $1/0$ as $\infty$).2012-11-07
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    Ah, makes sense. Thanks.2012-11-07