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I'm stuck with my homework:

Find both the radius and interval of convergence for the given series. Also, determine the values of $x$ for which the series converges absolutely and those for which the series converges conditionally.

(a) $\displaystyle\sum_{k=1}^{\infty}k(x-2)^k$

(b) $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{x}{2}\right)^k$

(c) $\displaystyle\sum_{k=1}^{\infty}k\left(-\frac{1}{3}\right)^k(x-2)^k$

(d) $\displaystyle\sum_{k=1}^{\infty}\frac{2^kk!}{k^k}x^k$

(e) $\displaystyle\sum_{k=1}^{\infty}\frac{k}{3^{2k-1}}(x-1)^{2k}$

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4  
Where are you stuck? –  joriki Dec 12 '11 at 9:40
    
I think I just need an example, I don't now how to do this correctly. –  Kevin Dec 12 '11 at 9:55
    
Here are some examples for the ratio test: en.wikipedia.org/wiki/Ratio_test#Examples –  joriki Dec 12 '11 at 10:18

1 Answer 1

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

3) The radius of convergence of the power series $\sum\limits_{n=1}^\infty a_n (x-a)^n$ is
$$ \tag{1} r=\lim_{n\rightarrow\infty} {|a_{n}|\over |a_{n+1}|}, $$ whenever this limit exists.

4) A power series converges absolutely for any $x$ in the interval $(a-r, a+r)$, and diverges for any $x$ with $|x-a|>r$, where $r$ is the radius of convergence.

Using these facts, if the radius of convergence is $r $:

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 $(-\infty,\infty)$ and the series converges (absolutely) for any $x$.

Case 2: If $r=0$, then the series converges only when $x=a$.

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

$$ (a-r, a+r), [a-r, a+r), (a-r, a+r], [a-r, a+r]. $$ The power series will converge absolutely for $x$ in $ (a-r, a+r)$ and will diverge for $|x-a|>r$. 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.

At the endpoints of the interval of convergence (if any), one may or may not have absolute convergence.

So, to solve one of your problems

1) Compute the radius of convergence, $R$, using $(1)$. The series will converge absolutely for any $x$ with $|x-a|<R$, and will diverge for any $x$ with $|x-a|>R$.

2) The first step deals with every $x$ except $x=a+R$ and $x=a-R$ (when $R$ is finite and non-zero).

You will need to examine the (normal) series $\sum_{n=1}^\infty a_n R^n $ and $\sum_{n=1}^\infty a_n(-1)^n R^n $ at this point. Each of these may or may not converge. If one does converge, it may be conditional. You have to go back to previous methods (comparison test, alternating series test, etc.) to determine the convergence class of each. Once you've done this, you can write down the interval of convergence and state whether you have conditional convergence, absolute convergence, or divergence at any endpoints of the interval.


Example:

Examine the convergence characteristics of the power series $\sum\limits_{n=0}^\infty 2^n x^n$.

We first compute $r$: $$ r=\lim_{n\rightarrow\infty}{|a_{n}|\over |a_{n+1}|} =\lim_{n\rightarrow\infty}{2^n\over 2^{n+1}} =\lim_{n\rightarrow\infty}{1\over 2}={1\over2}. $$ So, the radius of convergence is $r={1\over 2}$. At this point, we know that the series converges absolutely whenever $-{1\over2}<x<{1\over2}$, and diverges if $|x|>{1\over2}$. But we do not know what happens for $x={1\over2}$ or $x=-{1\over2}$.

We must examine these cases separately.

For $x=-{1\over2}$, we obtain the series $\sum\limits_{n=0}^\infty (-1)^n$. This series diverges by the $n^{\rm th}$-Term Test. So, $x=-{1\over2}$ is not in the interval of convergence.

For $x={1\over2}$, we obtain the series $\sum\limits_{n=0}^\infty 1$. This series diverges by the $n^{\rm th}$-Term Test. So, $x={1\over2}$ is not in the interval of convergence.

The interval of convergence is thus $( -{1\over2},{1\over2})$.

Moreover, we have shown:

1) The series converges absolutely for $-1/2 < x <1/2$.

2) The series diverges for $|x|\ge 1/2$.

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