# Calculus II

#### Marie1309

##### New member

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#### MarkFL

##### Super Moderator
Staff member
I suspect the issue is that the images are sideways, and not with your handwriting, which is actually very nice. If you're uploading images from a phone, then it's not really your fault...I see a lot of sideways images on forums from mobile users. Is this the question?

a) Find the radius and interval of convergence of:

$\sum_{n=0}^{\infty}\frac{n(3x-1)^n}{6^n}$

• topsquark

#### Marie1309

##### New member
Yes that’s correct.

#### HallsofIvy

##### Elite Member
Then use the "limit ratio test": $$\displaystyle \sum a_n$$ converges absolutely if $$\displaystyle \lim \left|\frac{a_{n+1}}{a_n}\right|< 1$$.

• MarkFL and topsquark

#### Marie1309

##### New member
This is how I started. Is it correct?

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#### Marie1309

##### New member
I suspect the issue is that the images are sideways, and not with your handwriting, which is actually very nice. If you're uploading images from a phone, then it's not really your fault...I see a lot of sideways images on forums from mobile users. Is this the question?

a) Find the radius and interval of convergence of:

$\sum_{n=0}^{\infty}\frac{n(3x-1)^n}{6^n}$
Yes that’s correct this is how I started but I’m not sure if it is correct Last edited by a moderator:

#### Marie1309

##### New member
Then use the "limit ratio test": $$\displaystyle \sum a_n$$ converges absolutely if $$\displaystyle \lim \left|\frac{a_{n+1}}{a_n}\right|< 1$$.
I did but I’m not sure if it is correct

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#### MarkFL

##### Super Moderator
Staff member
Yes that’s correct this is how I started but I’m not sure if it is correct

View attachment 11480
What we want is:

$$\displaystyle \frac{|3x-1|}{6}<1$$

or:

$$\displaystyle \left|x-\frac{1}{3}\right|<2$$

And so the radius of convergence is 2, and the interval of convergence is:

$$\displaystyle \left(\frac{1}{3}-2,\frac{1}{3}+2\right)$$

or:

$$\displaystyle \left(-\frac{5}{3},\frac{7}{3}\right)$$

• Marie1309 and topsquark

#### Jomo

##### Elite Member
Your work is very sloppy. When you replace n with n+1 you must write (n+1). The powers of (3x-1) are both n+1in your work. But this can't be since in one case n stays n and in the other case n becomes (n+1). You were told to take lim |{an+1/an|<1, yet you did not take the absolute value nor did you write limit. Also how did you compute that the radius of convergence is 6?

#### Marie1309

##### New member
Your work is very sloppy. When you replace n with n+1 you must write (n+1). The powers of (3x-1) are both n+1in your work. But this can't be since in one case n stays n and in the other case n becomes (n+1). You were told to take lim |{an+1/an|<1, yet you did not take the absolute value nor did you write limit. Also how did you compute that the radius of convergence is 6?
I think I got it now. So IS The radius of convergence 2?

#### Marie1309

##### New member
What we want is:

$$\displaystyle \frac{|3x-1|}{6}<1$$

or:

$$\displaystyle \left|x-\frac{1}{3}\right|<2$$

And so the radius of convergence is 2, and the interval of convergence is:

$$\displaystyle \left(\frac{1}{3}-2,\frac{1}{3}+2\right)$$

or:

$$\displaystyle \left(-\frac{5}{3},\frac{7}{3}\right)$$
Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?

#### MathProfessor

##### New member
If a power series converges for |x-a|<R and diverges for |x-a|>R, then the power series is centred at point a and has radius r=R. In your case a=1/3 and R=2. The interval of convergence is (a-R, a+R) = (-5/3,7/3) . The series diverges for both x=-5/3 and x=7/3 so the interval is open.

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#### Jomo

##### Elite Member
1st compute lim| an+1/an|

Set this lim <1
Then solve for x.

Example: |4x-6|<1
Factor out a 4
4|x-3/2| <1
|x - 3/2| < 1/4

So the interval of convergence is 1/4

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#### MarkFL

##### Super Moderator
Staff member
Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?
We have:

$$\displaystyle \frac{|3x-1|}{6}<1$$

And we want to get this in the form:

$$\displaystyle |x-a[<r$$

And so we factor out a 3 from the numerator and divide out common factors to get:

$$\displaystyle \frac{\left|x-\dfrac{1}{3}\right|}{2}<1$$

Multiply through by 2:

$$\displaystyle \left|x-\frac{1}{3}\right|<2$$

And so we have it in the form we want, where $$\dfrac{1}{3}$$ is the center of the interval of convergence, whose radius is $$2$$.

Does that make sense?

• Jomo

#### Marie1309

##### New member
We have:

$$\displaystyle \frac{|3x-1|}{6}<1$$

And we want to get this in the form:

$$\displaystyle |x-a[<r$$

And so we factor out a 3 from the numerator and divide out common factors to get:

$$\displaystyle \frac{\left|x-\dfrac{1}{3}\right|}{2}<1$$

Multiply through by 2:

$$\displaystyle \left|x-\frac{1}{3}\right|<2$$

And so we have it in the form we want, where $$\dfrac{1}{3}$$ is the center of the interval of convergence, whose radius is $$2$$.

Does that make sense?
RIGHTTT. Makes a lot of sense. Thank you so much for your time.

• MarkFL

#### Marie1309

##### New member
If a power series converges for |x-a|<R and diverges for |x-a|>R, then the power series is centred at point a and has radius r=R. In your case a=1/3 and R=2. The interval of convergence is (a-R, a+R) = (-5/3,7/3) . The series diverges for both x=-5/3 and x=7/3 so the interval is open.
Hey. Thanks for helping. But may I know why it diverges for both? . Thanks

#### MarkFL

##### Super Moderator
Staff member
Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
When we use the ratio test, we find the limit:

$$\displaystyle L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$

The ratio test states that:
• if $$L<1$$ then the series converges absolutely;
• if $$L>1$$ then the series is divergent;
• if $$L=1$$ or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
And so, because we require $$L<1$$, the interval of convergence is necessarily open, that is, we do not include the end-points.

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#### MathProfessor

##### New member
Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
When $$\displaystyle x=\frac{7}{3}$$ you have L=1 which is the case when the ratio test does not allow to conclude. In this case you replace x
by $$\displaystyle \frac{7}{3}$$ in $$\displaystyle a_n$$ and use the divergence test. Use the same step for $$\displaystyle x=\frac{-5}{3}$$. The interval of convergence is the set of points where the series converges.

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