# Power series: Is the radius of convergence $\frac{1}{3}$ for $\frac{x}{1\cdot3} + \frac{x^2}{2\cdot3^2} + \frac{x^3}{3\cdot3^3}+…$

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 $<1$.

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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

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.

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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 $|z| < 3.$

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