Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I should calculate the radius of convergenc and would like to know, if the result $\frac{1}{3}$ is correct.

Here the exercise:

$$ \frac{x}{1\cdot3} + \frac{x^2}{2\cdot3^2} + \frac{x^3}{3\cdot3^3} + \frac{x^4}{4\cdot3^4}... $$

This is: $$ \sum\limits_{n=0}^\infty \frac{x^{n+1}}{(n+1)3^{n+1}} \\ \lim\limits_{n \to \infty} \left| \frac{(n+1)\cdot3^{n+1}}{(n+2)\cdot3^{n+2}} \right| = \left| \frac{1}{3} \right| $$

I’m right? Thanks.

Summery

I could test with the ratio test if a power series is convergent. I could use $$\lim\limits_{n \to \infty} \frac{|a_{n+1}|}{\left|a_{n}\right|}$$ and get the $\left|x\right|$ for which the series is convergent. With that test the series is convergent, if the result is $<1$.

share|improve this question
3  
You have used the ratio test incorrectly, as far I can see. I think the series will converge for $|x| <3$ surely? –  Geoff Robinson Aug 28 '12 at 12:26
    
Remember that $$\frac{1}{R}=\limsup_{n \to +\infty} \sqrt[n]{|a_n|}.$$ Hence $R=3$. –  Siminore Aug 28 '12 at 12:27
    
Remember, the coefficients are $a_n=\frac{1}{(n+1)3^{n+1}}$ and you want $\lim_{n\to\infty} \frac{a_n}{a_{n+1}}$ –  Thomas Andrews Aug 28 '12 at 12:27
    
So I have to inverse the result? –  hofmeister Aug 28 '12 at 12:31

2 Answers 2

up vote 3 down vote accepted

Using the ratio test for absolute convergence.

$$ |a_{n+1}| = \frac{|x|^{n+2}}{(n+2)3^{n+2}} \\ $$

$$ |a_{n}| = \frac{|x|^{n+1}}{(n+1)3^{n+1}} \\ $$

$$ \frac{|a_{n+1}|}{\left|a_{n}\right|} = |x| \left( \frac{n+1}{n+2} \right)\left( \frac{1}{3} \right) $$

$$ \lim\limits_{n \to \infty} \frac{|a_{n+1}|}{\left|a_{n}\right|} = \frac{|x|}{3} $$

The series converges absolutely if $\frac{|x|}{3} < 1 $, which is when $|x| < 3$. Absolute convergence implies convergence.

You also need to check for convergence when $|x| = 3$ to determine if those points are in the radius of convergence.

share|improve this answer
    
Typo: the boundary is $|x|=3$, not $|x|=1$ :-) –  Siminore Aug 28 '12 at 12:46
    
Ok. I asked because I’m a little bit confused. In my formulary is written: $r = \lim\limits_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| $. In the exercises from the institute they use: $\lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $. I used the second way. I’m quite aware where’s the different. With the second way I get the value for which the $\left| x \right|$ the power series is convergence, right? –  hofmeister Aug 28 '12 at 12:47
    
@hofmeister Do you understand geometric series? A geometric series converges when the common ratio has absolute value less than 1. This is the basic idea of the ratio test. Take the ratio of term $|a_{n+1} / a_n|$. The series will converge if this ratio is less than 1. You will have an $x$ involved, so solve for $x$ to see for which $x$ values it converges. If you wanted to flip them upside down, you could, but then it would converge whenever the ratio is greater than 1. –  Graphth Aug 28 '12 at 12:55
    
@Siminore - Fixed my typo. Thanks for pointing it out! –  Legendre Aug 28 '12 at 12:57
    
@hofmeister: For $\sum a_n x^n$ (power series), your formulary is correct. For $\sum b_n$ (in which the $x$ stuff is incorporated in $b_n$, or your series is not a power series) approach described by Legendre is better. The two are equivalent for power series. You used an incorrect hybrid of the two approaches. –  André Nicolas Aug 28 '12 at 12:58

The easiest way to check the convergence of $a_0 + a_1(z-\omega) + a_2(z-\omega)^2 + \cdots$ is to apply the ratio test. To apply the ratio test, we need

$\lim_{k \to \infty} \frac{|a_{k+1}(z-\omega)^{k+1}|}{|a_k(z-\omega)^k|} < 1 \, .$

Simple algebraic manipulations show us that

$\lim_{k \to \infty} \frac{|a_{k+1}(z-\omega)^{k+1}|}{|a_k(z-\omega)^k|} < 1 \iff |z-\omega| < \lim_{k \to \infty}\left| \frac{a_k}{a_{k+1}} \right| . $

In other words, if we denote the far right hand limit (assuming it exists) by $\rho$, then the ratio test tells us that the series converges for all $z$ within a distance of $\rho$ of $\omega,$ i.e. $|z-\omega| < \rho.$

I think your problem was that you had the $a_k$ and $a_{k+1}$ the wrong way up and you got the reciprocal of the radius of convergence. In your case $\omega = 0$ and $a_k = 1/(k+1)3^{k+1}$, as you have correctly identified. It follows that

$\rho = \lim_{k\to\infty} \left| \frac{(k+2)3^{k+2}}{(k+1)3^{k+1}} \right| = 3\lim_{k \to \infty} \left|\frac{k+2}{k+1} \right| = 3 \, .$

Thus, your series converges for all $|z| < 3.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.