Calculus II

Marie1309

New member
Hello, could someone help me doing this exercice please? I’m really struggling.

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Marie1309

New member
Is it clearer now?

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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:

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

Marie1309

New member
Yes that’s correct.

HallsofIvy

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

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

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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:

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

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?

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.

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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