# Calculus: Interval of Convergence of Power Series

Here is the problem:

Find the interval of convergence of the power series $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(x-2)^{n}}{n4^{n}}$

So far I have tried the Ratio Test which got me $\frac{x-2}{4}$, but I have no idea what to do with this kind of outcome. Did I do something wrong?

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So you want the absolute value of the ratio to be smaller than $1$, right? Your task is to determine what this imposes on $x$... – Did Dec 14 '11 at 22:26
Hints: 1) Ratio Test tells the series $\sum a_n$ converges if $|\lim_n \frac{a_{n+1}}{a_n}|<1$; hence for your series to converge it is sufficient that $|\frac{x-2}{4}|<1$; now solve for $x$... 2) On the other hand, the interval of convergence cannot be bigger than the one determined in 1: in fact, does $\sum_{n=1} (-1)^{n-1} \frac{(x-2)^n}{n4^n}$ converges if $x=-2$? – Pacciu Dec 14 '11 at 22:31
@Didier Oh okay, so I guess X has to be -2 < X < 4? Is that correct? – Kyle V. Dec 14 '11 at 22:33
@StickFigs Actually, since Ratio Test gives only a sufficient condition for a series to converge, you cannot be sure that $]-2,6[$ is the largest interval in which your series converges; in other words, theorically your series could converge into a larger open interval $J\supset ]-2,6[$. This latter case cannot occur: in fact your series diverges in $x=-2$ and in $x=6+\varepsilon$ ($\varepsilon >0$ arbitrarily small), thus it cannot converge into an interval $J$ strictly larger than $]-2,6[$ for known results on power series. – Pacciu Dec 14 '11 at 22:49
@Pacciu : The ratio test says if the limiting ratio of the absolute values of successive terms is less than $1$, then the series converges (so that's a sufficient condition for convergence) and if the limit is more than $1$, then the series diverges (so that's a necessary condition for convergence). (Of course, some series converge and have a limiting ratio of exactly $1$, and some others diverge and have a limiting ratio of exactly $1$.) – Michael Hardy Dec 15 '11 at 4:47

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

That the interval of convergence is, indeed, an interval follows from the following facts:

1) Suppose the power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$ converges
for $x=b$. Set $|b-a|=r$. Then the power series converges absolutely for all $x$ with $|x-a|<r$.

2) Suppose the power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$ diverges for $x=b$. Set $|b-a|=r$. Then the power series diverges for all $x$ with $|x-a|>r$.

I will emphasise here: the interval of convergence of a power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$ is an interval centered at $a$

Luckily, we can find "most" of the interval of convergence by first finding the radius of convergence, $R$, of the power series. $R$ is defined to be half the length of the interval of convergence (if the interval of convergence is all of $\Bbb R$, we set $R=\infty$), and may be computed using

$$\tag{1} R=\lim_{n\rightarrow\infty} {|a_{n}|\over |a_{n+1}|},$$ provided this limit exists (including the case where the limit is $\infty$).

Suppose we found the radius of convergence, $R$ of a power series. Then we would know, from facts 1) and 2), that the power series converges for any $x$ in the interval $(a-R, a+R)$, and diverges for any $x$ with $|x-a|>R$. Thus the interval of convergence is at least the set $(a-r, a+r)\cup\{a\}$ and at most that set together with one or both of its endpoints.

Finding the interval of convergence once the radius of convergence is known is broken down into three cases:

Case 1: If $R$ is infinite, then the interval of convergence is $(-\infty,\infty)$.

Case 2: If $R=0$, then the series converges only when $x=a$, and the interval of convergence is $\{a\}$.

Case 3: When $R$ is a finite non-zero number, the interval of convergence will be one of

$$(a-R, a+R), [a-R, a+R), (a-R, a+R], [a-R, a+R].$$

To determine which of the four intervals above is the interval of convergence, one must examine the series obtained when $x$ is replaced by $a+R$ and $a-R$ separately.

So, to find the interval of convergence of the power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$

1) Compute the radius of convergence, $R$, using $(1)$ (if this limit does not exist, you might try computing $R$ by using the formula given by the Root test).

2) If $R=0$ the interval of convergence is $\{a\}$.

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

3) If $R$ is finite and non-zero, the interval of convergence is one of

$$\tag{2} (a-R, a+R), [a-R, a+R), (a-R, a+R], [a-R, a+R].$$

Examine the series obtained when $x=a+R$ and $x=a-R$. These will give the series $$\sum_{n=0}^\infty a_n R^n, \text{ and }\quad\sum_{n=0}^\infty a_n(-1)^n R^n .$$ Each of these may or may not converge. You have to go back to previous methods (comparison test, alternating series test, etc.) to determine whether these series converge or not. Once you've done this, you can determine which of the four intervals in (2) is the interval of convergence.

Example:

Find the interval of convergence of the power series $\sum\limits_{n=0}^\infty {(-1)^{n-1}\over n4^n } (x-2)^n$.

We first compute $R$: \eqalign{ R&=\lim_{n\rightarrow\infty}{|a_{n}|\over |a_{n+1}|} \cr &=\lim_{n\rightarrow\infty}{{1\over n4^n}\over{1\over (n+1)4^{n+1}}} \cr &=\lim_{n\rightarrow\infty}{{1\over n4^n}\cdot{ (n+1)4^{n+1}}} \cr &=\lim_{n\rightarrow\infty}{4(n+1)\over n}={4}. } So, the radius of convergence is $R=4$. At this point, we know that the
interval of convergence is one of (using $a= 2$): $$[-2,6], \quad (-2,6),\quad [-2,6), \quad (-2,6].$$
We do not know at this point what happens for $x={-2}$ or $x=-{6}$.

We must examine these cases separately.

For $x=-{ 2}$, we obtain the series $$\sum\limits_{n=1}^\infty (-1)^{n-1} {(-2-2)^n\over n 4^n}=\sum_{n=1}^\infty\,- 1 .$$ This series diverges by the $n^{\rm th}$-Term Test. So, $x=-2$ is not in the interval of convergence.

For $x={6}$, we obtain the series $$\sum\limits_{n=1}^\infty (-1)^{n-1} {(6-2)^n\over n 4^n}=\sum_{n=1}^\infty\,(-1)^{n-1} {1\over n } .$$ This series is a convergent Alternating series. So $x=6$ is in the interval of convergence.

The interval of convergence is thus $(-2,6]$.

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