Finding radius of convergence for a infinite series?

**MathIsOhSoHard** · October 22nd 2012, 07:17 PM

Find the radius of convergence for the following infinite series:
$\sum_{n=1}^\infty\frac{x^{2n}}{3^n(n^2+1)}$

Answer: $\sqrt{3}$

Use the following formula:
$\left|\frac{a_{n+1}}{a_n}\right|$

My attempt:
$\left|\frac{x^{2(n+1)}}{3^{(n+1)}((n+1)^2+1)}\cdot \frac{3^n(n^2+1)}{x^{2n}}\right|=\left|\frac{x^{2( n+1)}}{x^{2n}}\cdot\frac{3^n(n^2+1)}{3^{(n+1)}((n+ 1)^2+1)}\right|=|x|\frac{3^n(n^2+1)}{3^{(n+1)}((n+ 1)^2+1)}$

I'm not sure if I'm on the right path. I can't seem to figure it out what to do.
I know that for convergence we need that $|x|<r$ and for divergence $|x|>r$ .

Any help?

**Prove It** · October 22nd 2012, 07:19 PM

Originally Posted by MathIsOhSoHard

Find the radius of convergence for the following infinite series:
$\sum_{n=1}^\infty\frac{x^{2n}}{3^n(n^2+1)}$

Answer: $\sqrt{3}$

Use the following formula:
$\left|\frac{a_{n+1}}{a_n}\right|$

My attempt:
$\left|\frac{x^{2(n+1)}}{3^{(n+1)}((n+1)^2+1)}\cdot \frac{3^n(n^2+1)}{x^{2n}}\right|=\left|\frac{x^{2( n+1)}}{x^{2n}}\cdot\frac{3^n(n^2+1)}{3^{(n+1)}((n+ 1)^2+1)}\right|=|x|\cdot\frac{3^n(n^2+1)}{3^{(n+1) }((n+1)^2+1)}$

I'm not sure if I'm on the right path. I can't seem to figure it out what to do.
I know that for convergence we need that $|x|<r$ and for divergence $|x|>r$ .

Any help?

Every series that satisfies $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}$ is convergent.

So once you get an expression for $\displaystyle \begin{align*} \left| \frac{a_{n + 1}}{a_n} \right| \end{align*}$ , which will be in terms of $\displaystyle \begin{align*} x \end{align*}$ , evaluate the limit, then set this limit value less than 1 and solve for $\displaystyle \begin{align*} x \end{align*}$ .

**MathIsOhSoHard** · October 22nd 2012, 07:29 PM

Originally Posted by Prove It

Every series that satisfies $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}$ is convergent.

So once you get an expression for $\displaystyle \begin{align*} \left| \frac{a_{n + 1}}{a_n} \right| \end{align*}$ , which will be in terms of $\displaystyle \begin{align*} x \end{align*}$ , evaluate the limit, then set this limit value less than 1 and solve for $\displaystyle \begin{align*} x \end{align*}$ .

Do you want me to evaluate $\frac{3^n(n^2+1)}{3^{(n+1)}((n+1)^2+1)}$ ?

Simplified, it becomes:
$\frac{3^n(n^2+1)}{3^{(n+1)}((n+1)^2+1)}=\frac{1}{3 }\frac{n^2+1}{n^2+2n+2}$

If I evaluate its limit for $n\rightarrow\infty$ :

$\lim_{n\rightarrow\infty}\frac{1}{3}\frac{n^2+1}{n ^2+2n+2}=\frac{1}{3}$

Is this what you meant?

**Prove It** · October 22nd 2012, 07:42 PM

First of all, you did not evaluate that limit correctly. Second, that's not the limit you are trying to evaluate.

You tried to evaluate $\displaystyle \begin{align*} \lim_{n \to \infty}\left| \frac{3^n\left( n^2 + 1 \right)}{3^{n+1}\left[ (n + 1)^2 + 1 \right]} \right| \end{align*}$ when you should be evaluating $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{x^{2(n+1)}}{x^{2n}} \cdot \frac{3^n \left( n^2 + 1 \right)}{3^{n+1} \left[ (n+1)^2 + 1 \right]} \right| \end{align*}$ . Luckily this simplifies to $\displaystyle \begin{align*} |x|^2 \lim_{n \to \infty}\left| \frac{3^n \left( n^2 + 1 \right)}{3^{n+1} \left[ (n + 1)^2 + 1 \right]} \right| \end{align*}$ so fixing this shouldn't be too much trouble.

Then, like I said, once you have evaluated this limit, set it less than 1 and solve for $\displaystyle \begin{align*} x \end{align*}$ .

**MathIsOhSoHard** · October 22nd 2012, 07:50 PM

Originally Posted by Prove It

First of all, you did not evaluate that limit correctly. Second, that's not the limit you are trying to evaluate.

You tried to evaluate $\displaystyle \begin{align*} \lim_{n \to \infty}\left| \frac{3^n\left( n^2 + 1 \right)}{3^{n+1}\left[ (n + 1)^2 + 1 \right]} \right| \end{align*}$ when you should be evaluating $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{x^{2(n+1)}}{x^{2n}} \cdot \frac{3^n \left( n^2 + 1 \right)}{3^{n+1} \left[ (n+1)^2 + 1 \right]} \right| \end{align*}$ . Luckily this simplifies to $\displaystyle \begin{align*} |x|^2 \lim_{n \to \infty}\left| \frac{3^n \left( n^2 + 1 \right)}{3^{n+1} \left[ (n + 1)^2 + 1 \right]} \right| \end{align*}$ so fixing this shouldn't be too much trouble.

Then, like I said, once you have evaluated this limit, set it less than 1 and solve for $\displaystyle \begin{align*} x \end{align*}$ .

Ah yes, my mistake.

So we now get that:
$\lim_{n\rightarrow\infty}x^2\frac{1}{3}\frac{n^2+1 }{n^2+2n+2}=\frac{1}{3}x^2$

Solving x for:
$\frac{1}{3}x^2=1$

$x=\sqrt{3}$

So since it is solved for less than 1, this is pretty much the radius where every value inside the circle is convergence. I think I get it now, thanks!

**Prove It** · October 22nd 2012, 07:52 PM

Originally Posted by MathIsOhSoHard

Ah yes, my mistake.

So we now get that:
$\lim_{n\rightarrow\infty}x^2\frac{1}{3}\frac{n^2+1 }{n^2+2n+2}=\frac{1}{3}x^2$

Solving x for:
$\frac{1}{3}x^2=1$

$x=\sqrt{3}$

So since it is solved for less than 1, this is pretty much the radius where every value inside the circle is convergence. I think I get it now, thanks!

Do NOT forget your absolute value signs! You should get $\displaystyle \begin{align*} |x| < \sqrt{3} \end{align*}$ , which in the real numbers is an interval between $\displaystyle \begin{align*} -\sqrt{3} \end{align*}$ and $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ and in the complex number is a circle centred at the origin of radius $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ . You should not get $\displaystyle \begin{align*} x < \sqrt{3} \end{align*}$ which is an infinitely long interval of all numbers less than $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ .

**MathIsOhSoHard** · October 22nd 2012, 08:07 PM

Originally Posted by Prove It

Do NOT forget your absolute value signs! You should get $\displaystyle \begin{align*} |x| < \sqrt{3} \end{align*}$ , which in the real numbers is an interval between $\displaystyle \begin{align*} -\sqrt{3} \end{align*}$ and $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ and in the complex number is a circle centred at the origin of radius $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ . You should not get $\displaystyle \begin{align*} x < \sqrt{3} \end{align*}$ which is an infinitely long interval of all numbers less than $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ .

Yes, I see.
So if I want to show that the series is convergent for $x=\pm\sqrt{3}$ then all I got to do is insert a value x that is smaller than $\sqrt{3}$ in the series, right?
For example:
$\sum_{n=1}^\infty\frac{\sqrt{2}^{2n}}{3^n(n^2+1)}$

This gives a value. However if a value bigger than $\sqrt{3}$ is inserted, such as $\sqrt{4}$ then the series approaches infinite and so it's divergent.
This fits in accordance with the circle where any value less than the radius is inside the circle and thus convergence, and any value bigger than the radius is outside the circle and divergence. Did I understand it correctly?

**Prove It** · October 22nd 2012, 08:22 PM

The series is divergent for $\displaystyle \begin{align*} x < -\sqrt{3} \end{align*}$ or $\displaystyle \begin{align*} x > \sqrt{3} \end{align*}$ from the ratio test, because the limit ends up greater than 1, there is no need to check them.

The only values you will have to check are the values such that the limit actually equals 1, since that is when the ratio test is inconclusive. So to check the convergence that the endpoints $\displaystyle \begin{align*} \pm \sqrt{3} \end{align*}$ , substitute these values in and use a different test on each of the resulting series.

Having said that, in this question you don't need to check them, because you are only asked for the RADIUS of convergence. You have already found that it is $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ .

**MathIsOhSoHard** · October 23rd 2012, 12:32 PM

Originally Posted by Prove It

The series is divergent for $\displaystyle \begin{align*} x < -\sqrt{3} \end{align*}$ or $\displaystyle \begin{align*} x > \sqrt{3} \end{align*}$ from the ratio test, because the limit ends up greater than 1, there is no need to check them.

The only values you will have to check are the values such that the limit actually equals 1, since that is when the ratio test is inconclusive. So to check the convergence that the endpoints $\displaystyle \begin{align*} \pm \sqrt{3} \end{align*}$ , substitute these values in and use a different test on each of the resulting series.

Having said that, in this question you don't need to check them, because you are only asked for the RADIUS of convergence. You have already found that it is $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ .

There are a couple of other exercises (including this one) that asks me to check the convergence for the endpoints.
How would I go about doing so? Do I just insert the endpoints in the series and then use tests or is there another way? My book doesn't explain anything about the endpoints, it only mentions scenarios where radius is smaller or bigger, but never equal.

**Prove It** · October 23rd 2012, 04:08 PM

Originally Posted by MathIsOhSoHard

There are a couple of other exercises (including this one) that asks me to check the convergence for the endpoints.
How would I go about doing so? Do I just insert the endpoints in the series and then use tests or is there another way? My book doesn't explain anything about the endpoints, it only mentions scenarios where radius is smaller or bigger, but never equal.

Yes that is exactly what you do. Substitute in the endpoints and use an appropriate test for convergence. Obviously the ratio test won't work, but the comparison test, root test or integral test might.

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