Finding the Radius of Convergence for $zsin(z^2)$ in Maclaurin Series

[MissP.25_5]

Hello.
I need explanation on why the answer for this problem is R = ∞.

Here's the question and the solution.

Expand the function into maclaurin series and find the radius of convergence.
$zsin(z^2)$

Solution:
$$zsin(z^2)=z\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

Divide both sides by z,

$$sin(z^2)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

So here's the calculation but I don't know how to get the radius of convergence. Answer is ∞.

 

[Mogarrr]

Try the ratio test.

Need a refresher? I did. Look at Paul's Online Calculus notes: http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx.

 

[MissP.25_5]

Try the ratio test.

Need a refresher? I did. Look at Paul's Online Calculus notes: http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx.

I did the ratio test and got 4.

 

[Mogarrr]

I think the ratio test should start off like this

$lim_{n \to \infty} | \frac {(-1)^{n+1} \cdot z^{2 \cdot (2(n+1)+1)})}{(2(n+1)+1)!} \cdot \frac {(2n+1)!}{(-1)^{n} \cdot z^{2 \cdot (2n+1)}} |$

You should find that the limit is 0, so R = ∞.

 

[MissP.25_5]

I think the ratio test should start off like this

$lim_{n \to \infty} | \frac {(-1)^{n+1} \cdot z^{2 \cdot (2(n+1)+1)})}{(2(n+1)+1)!} \cdot \frac {(2n+1)!}{(-1)^{n} \cdot z^{2 \cdot (2n+1)}} |$

You should find that the limit is 0, so R = ∞.

That's what I did, only that I use the formula to find R directly, instead of L because R is just 1/L.
Could you show me the steps until you get 0, please? Because as you see mine, I got 4 :/

 

[MissP.25_5]

I think the ratio test should start off like this

$lim_{n \to \infty} | \frac {(-1)^{n+1} \cdot z^{2 \cdot (2(n+1)+1)})}{(2(n+1)+1)!} \cdot \frac {(2n+1)!}{(-1)^{n} \cdot z^{2 \cdot (2n+1)}} |$

You should find that the limit is 0, so R = ∞.

Oh, I get it now! I just googled about limits and learned it again. I think I understand now why the lmit is infinity. Thanks!

 

[HallsofIvy]

For others who might be wondering here is your mistake: near the end you have
$$\lim_{n\to\infty}\left|-4n^2+ 10n+ 6\right|$$
and then immediately have
$$\lim_{n\to\infty}\left|-(4- \frac{10}{n}- \frac{6}{n^2})\right|$$
where you have divided by n^2. Obviously you cannot do that! You have essentially factored out "$n^2$" and should have
$$\left(\lim_{n\to\infty}n^2\right)\left(\lim_{n\to\infty}\left|4- \frac{10}{n}- \frac{6}{n^2}\right|\right)$$

The limit on the right is 4 but the limit on the left is $\infty$. Their product is "$\infty$".

 

[Mogarrr]

That's what I did, only that I use the formula to find R directly, instead of L because R is just 1/L.
Could you show me the steps until you get 0, please? Because as you see mine, I got 4 :/

I think there's some confusion about the ratio test. For a series $\sum a_n$, the ratio test is $lim_{n \to \infty} \frac {a_{n+1}}{a_n}$.

This is exactly what I started off with when I saw that you had simplified the series. I start off with

$lim_{n \to \infty} | \frac {(-1)^{n+1} \cdot z^{2 \cdot (2(n+1)+1)})}{(2(n+1)+1)!} \cdot \frac {(2n+1)!}{(-1)^{n} \cdot z^{2 \cdot (2n+1)}} |$

Since this is an absolute value, the $-1$ terms can be ignored, and simplifying the exponents of the first fraction, you should see

$lim_{n \to \infty} | \frac {z^{4n+6}}{(2n+3)!} \cdot \frac {(2n+1)!}{z^{4n+2}} |$

To simplify, observe that $(2n+3)! = (2n+3) \cdot (2n+2) \cdot (2n+1)!$, $z^{4n+6} = z^{4n} \cdot z^6$ and that $z^{4n+2} = z^{4n} \cdot z^2$.

So we have

$lim_{n \to \infty} | \frac {z^{4n} \cdot z^6}{(2n+3) \cdot (2n+2) \cdot (2n+1)!} \cdot \frac {(2n+1)!}{z^{4n} \cdot z^2} |$

The $z^{4n}$'s and $(2n+1)!$'s cancel out. You can also factor the remaining z terms. This will leave you with

$\frac {z^6}{z^2} lim_{n \to \infty} \frac 1{(2n+3)(2n+2)}$.

The limit is 0. Interpret this to mean, it does not matter what values you choose for the variable z. Any real number will work. So we have $-\infty < z < \infty$, so the radius of convergence is ∞.

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