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Problem:

$\sum\limits_{k=0}^\infty(k+2)^2\cdot\frac{(3x+1)^k}{2^k\cdot2^2}$

Determine whether this is convergent or divergent. Use the Radius of Convergence

Hey there,

I ran into this Exercise and found myself a little bit of trouble. According to my teacher, before I can use the Radius of Convergence, I need to put this equation into a form like:

$\sum\limits_{k=0}^\infty a_n\cdot x^k$

I'm a bit embarassed but I can't transform this. I just follow up with my thoughts and how far I went.

$\sum\limits_{k=0}^\infty(k+2)^2\cdot\frac{(3x+1)^k}{2^k\cdot2^2}$

= $\sum\limits_{k=0}^\infty(k^2+4k+4)\cdot\frac{(3x+1)^k}{2^k\cdot4}$ = $\sum\limits_{k=0}^\infty\frac{(k+2)^2}{4}\cdot\frac{(3x+1)^k}{2^k}$

Now I would do $\sqrt[k]{\frac{(3x+1)^k}{2^k}}$ But that would eliminate my $x^k$

Kind of running circles here. Any help is appreciated

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2 Answers 2

up vote 2 down vote accepted

You’re taking your teacher’s statement a little too literally. What it really means is that you need to get it in the form $\sum\limits_{k\ge 0}a_ky^k$, where the coefficients $a_k$ don’t depend on $x$, and $y$ does. In this problem you want to rewrite it as follows:

$$\sum_{k=0}^\infty(k+2)^2\cdot\frac{(3x+1)^k}{2^k\cdot2^2}\sum_{k\ge 0}\left(\frac{k+2}2\right)^2\left(\frac{3x+1}2\right)^k\;.$$

Here $a_k=\left(\dfrac{k+2}2\right)^2$ and $y=\dfrac{3x+1}2$. When you apply the ratio test, your $(k+1)$-st term will be

$$\left(\frac{(k+1)+2}2\right)^2\left(\frac{3x+1}2\right)^{k+1}\;.$$

You’ll end up getting a bound on $|y|$, i.e., on $\left|\dfrac{3x+1}2\right|$, something like $$\left|\frac{3x+1}2\right|<c\;;$$ solve that inequality for $x$, and you’ll get something of the form $a<x<b$, so that the radius of convergence is $\dfrac{b-a}2$, half the length of the interval of convergence.

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Thank you so much Brian! Seems like I was too narrowminded with this :( –  blacksmth Nov 30 '12 at 8:20
    
@blacksmth: You’re very welcome. Maybe not narrow-minded; you may simply not have seen an example of this type to help you to realize what the teacher was actually saying. –  Brian M. Scott Nov 30 '12 at 8:23

First recall that $$\sum_{n=1}^{\infty} a_n$$ converges if $$\limsup_{n} \vert a_n \vert^{1/n} < 1$$ In your case, your $a_n = (n+2)^2 \dfrac{(3x+1)^n}{2^n \cdot 2^2}$.

Note that $$\left \vert \dfrac{(n+2)^2}{2^2}\right \vert^{1/n} \to 1 \,\,\,\,\,\,\,\,(\text{Why?})$$

Hence, $$\left \vert (n+2)^2 \dfrac{(3x+1)^n}{2^n \cdot 2^2}\right \vert^{1/n} < 1 \implies \left \vert \dfrac{3x+1}2\right \vert < 1 \implies -2 < 3x+1 < 2$$ Hence, $$-1 < x < \dfrac13$$ is your desired interval of convergence.

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Thank you for your answer Marvis! The Why is a bit confusing. But i'm trying to get behind it! –  blacksmth Nov 30 '12 at 8:22
    
$$\left \vert \dfrac{(n+2)^2}{2^2}\right \vert^{1/n} \to 1 \,\,\,\,\,\,\,\,(\text{Why?})$$ I don't get why my $a_n$ is 1? –  blacksmth Nov 30 '12 at 8:38
    
@blacksmth Are you aware of the following limit $\lim_{n \to \infty} n^{1/n} = 1$? –  user17762 Nov 30 '12 at 8:41
    
Ah,yes. Now I see it. Sorry!! –  blacksmth Nov 30 '12 at 8:45

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