Sum from 0 to infinity of [3/[(n+1)(5^n)]](x-3)^n HELP

[Lo.Lee.Ta.]

Sum from 0 to infinity of [3/[(n+1)(5^n)]](x-3)^n HELP :)

1.
∞
Ʃ $\frac{3}{(n+1)(5^{n}}$*(x-3)ⁿ
n=0


2. The first question I had to answer was: What is f(3)?

I found the first 4 terms to be:

3, 3/10(x-3), 3/75(x-3)², 3/189(x-3)³

So f(3) equals 3, I'm pretty sure.
Because all the other (except the 1st) cancel to equal 0.
3 + infinity zeros = 3


The second question asks me whether or not f(4) converges.


Well, that would be: 3 + (3/10)(1)¹ + (3/75)(1)²+(3/189)(1)³ + ...

3 + increasingly small numbers = convergence

Is this the right way to think about it? Is this is correct, is there a better, perhaps faster way?




The third question asks whether or not f(8) converges.


That would be: 3 + (3/10)(5)¹ + (3/75)(5)² + (3/189)(5)³ + ...

= 3 + increasingly small numbers = convergence

...I don't think it's getting bigger very fast... It seems like it would take way too long for this to go to infinity. That's why I'm thinking it would converge...



The fourth question asks whether or not f(-2) converges.

That would be: 3 + (3/10)(5)¹ + (3/75)(5)² + ...

Yeah, this is the same as last time. So I say it converges again.



Now, the fifth question asks what the interval of convergence is for this power series.

To find this interval, I think I need to use the formula: a_n/a_n+1

$\frac{3(x-3)^{n}}{(n+1)5^{n}}$*$\frac{(n+2)(5^{n+1})}{3(x-3)^{n+1}}$

= $\frac{5(n+2)}{(n+1)(x-3)}$

Now, the next step is to take the limit, but in this form, I'll get ∞/∞.
I could do L'Hospital's Rule, but I think it might be easier if I rewrote the problem before I take the limit.

Hopefully, I'm not doing any illegal math moves... :/

$\frac{5}{(n+2)^{-1}(n+1)(x-3)}$

=$\frac{5}{n^{0}+n^{-1}+2n^{-1}+2{-1}(x-3)}$

= $\frac{5}{1 + 1/n + 1/2n + 1/2(x-3)}$

lim |$\frac{5}{1.5 + 1/n + 1/2n(x-3)}$|
n→∞

= |5/[(1.5 + 0 + 0)(x-3)]|

I CAN'T FIND THE GREATER THAN SYMBOL, so I'm using "∠"!

= -1 ∠ 10/3(x-3) ∠ 1

= -33/10 ∠ x ∠ 33/10

...So it is convergent between this interval... Which I'm pretty sure is WRONG!

That is too hideous of an answer to be correct. AND- it does not fit with my convergence answers from before!
33/10= 3.3

4 AND 8 lie outside of this interval. Something's wrong here... :/

Please help! D=

Thanks :)

 

[STEMucator]

Err let's start at the top.

You said f(3) = 3? Check that again.

 

[Lo.Lee.Ta.]

Oh, yeah. Seems like the first term should really be 0. (0)⁰ I guess is 0? My calculator says error when I put that in. But should I say that's 0 or undefined or what?

Let me just write everything out.

$\frac{3}{(0+1)(5^{0})}$*(3-3)⁰ + $\frac{3}{(1+1)(5^{1})}$*(3-3)¹ + $\frac{3}{(2+1)(5^{2})}$*(3-3)²+ $\frac{3}{(3+1)(5^{3})}$*(3-3)³

 

[Infrared]

In this case 0^0 is 1, not zero since when n=0, you get your constant term. This should not be affected by your choice of x.
It isn't the best notation maybe but this is the case for all power series that have a constant term.

 

[Mark44]

1.
∞
Ʃ $\frac{3}{(n+1)(5^{n}}$*(x-3)ⁿ
n=0


The second question asks me whether or not f(4) converges.


Well, that would be: 3 + (3/10)(1)¹ + (3/75)(1)²+(3/189)(1)³ + ...

3 + increasingly small numbers = convergence

Is this the right way to think about it? Is this is correct, is there a better, perhaps faster way?

This is definitely NOT the right way to think about this. Using this logic, you would conclude that this series converges:
$$ \sum_{n = 1}^{\infty}\frac{1}{n} = 1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ...$$
Here you have 1 + "increasingly small numbers", but it's well known that this series (the harmonic series) diverges.

For this part of the problem, you have
$$f(4) = \sum_{n = 0}^{\infty}\frac{3}{(n + 1)5^n}1^n$$
Since 1ⁿ = 1 for each finite value of n, the above can be simplified to
$$f(4) = \sum_{n = 0}^{\infty}\frac{3}{(n + 1)5^n}$$
I'm sure you've worked with series of constants before now, so what theorems can you use to determine whether this series converges or diverges?

I CAN'T FIND THE GREATER THAN SYMBOL, so I'm using "∠"!

Most keyboards have the > symbol on the same key as the comma. Press SHIFT + , to get the > symbol.

Also, if you click Go Advanced on the lower right below the PF text entry area, an advanced menu bar opens across the top of the window, and a table of quick symbols appears at the right. This table as a number of symbols, including ≤, ≥ (which aren't usually available on keyboards), ≠, and other symbols.

 

[Lo.Lee.Ta.]

∞
Ʃ $\frac{3}{(n+1)(5^{n})}$*(x-3)ⁿ
n=0


Since n's are in the exponents, I would use the ratio/root test.
And this is a geometric series, right...?

Is it that the only way I can figure this out is to find the interval of convergence?

Once I find the interval of convergence, I can just see if f(4), f(8), etc. lie within that interval and converge?

Well, I did that below. I'm not sure if the answer is right... :/

Since I found the interval to be between -3.3 and 3.3, then I would say that f(3) converges, f(4) diverges, f(8) diverges, and f(-2) converges?

Thanks! :)

 

[Infrared]

It is impossible for the convergence set to get (-3.3,3.3) with one or both of the endpoints possible added. There is a nice theorem that for any power series, the convergence set is centered at the center of the power series. (-3.3,3.3) is centered at zero while the power series is centered at 3. I got something else. (Hint: there is a reason why the question is asking about x=-2 and x=8).

Edit: Also, no, it is not geometric.

 

[Lo.Lee.Ta.]

I just followed the a_n/a_n+1 formula to arriave at my answer...

Do you see anything wrong with my work that gave me the wrong interval of -3.3 to 3.3?

I don't see what's wrong here... :/

 

[Infrared]

The ratio test considers $|\frac{a_{n+1}}{a_n}|$ as n becomes large (not the other way around) . Where the this is greater than one it diverges, less than one it converges, and inconclusive if it equals one. So when is $\lim_{n\rightarrow \infty} |{\frac{(x-3)(n+1)}{5(n+2)}}|$ less than one? What happens when it equals one?

 

[Lo.Lee.Ta.]

Wow, that was dumb of me. O_O It's "a_n+1/a_n"! ugh.

okay, so now I have: $\frac{(x-3)(n+1)}{5(n+2)}$

Then I was thinking that maybe I should multiply the numerator and the denominator by 1/n because then I'll have:

lim $\frac{(x-3)(1 + 1/n)}{5(1 + 2/n)}$
n→∞


Now, there's no question that when I put in ∞ for n, the term will go to zero.

That leaves me with:

|$\frac{(x-3)}{5}$|< 1

-1 < $\frac{x-3}{5}$ < 1

-2 < x < 8

Hey! That's like the question! haha
So, I guess since the interval is only GREATER THAN -2 and LESS THAN 8, then f(-2) and f(8) diverge.

Is this right?

f(3) converges, f(4) converges, f(8) diverges, and f(-2) diverges.

Is it possible to have an interval between non-whole numbers?

If the expression equals 1, it's inconclusive; we have to do another test (integral test).

If the expression is greater than 1, it fails the test and is divergent, right?

Thanks! :)

 

[Infrared]

How did you get that the series diverges for x=-2 and x=8? The test is inconclusive when the limit of the ratio is one, so you have to check these points manually. Yes, it is of course possible for the convergence set to be between two non-integers. What happens if you replace the 5 by a 5.1?

 

[Lo.Lee.Ta.]

For my expression: |$\frac{(x-3)}{5}$| < 1

I'm trying to find for which values of x is my expression convergent (or less than 1).

-1 < $\frac{(x-3)}{5}$ < 1

-5 < x-3 < 5

-2 < x < 8

So, it says that when the x value is between -2 and 8, my expression is convergent.
Outside of that interval, it is divergent.

So, f(-2) is outside the interval, so I say it's divergent, etc.

...I don't really understand what you mean about having to check points manually... :/

I thought that when you have an expression with only 1 variable (not like the power series which has the a_n and the x), then you'd end up with a number at the end of the ratio test.

Then you see if that number is less than 1. If it is, then the entire expression converges.
If it isn't, the entire series diverges.
If it equals one, you've got to do the integral test.

Hmmm. :/ I thought that was the whole point of finding the interval- to know what x values make our expression converge...

Help. Thanks! :)

 

[Infrared]

No, the ratio test works like this: If the ratio is less than one, it definitely converges,
If the ratio is equal to one, then the test is INCONCLUSIVE, if greater than one then it diverges.

 

[Lo.Lee.Ta.]

Okay, so you say I should test each x value manually.

|$\frac{(x-3)}{5}$| < 1

f(3)= |$\frac{3-3}{5}$ < 1

$\frac{0}{5}$ = 0 < 1

This would converge.

f(4)= |$\frac{(4-3)}{5}$|

= |$\frac{1}{5}$| < 1

This would converge.

f(8)= |$\frac{8-3}{5}$|

= |$\frac{5}{5}$| = 1 is NOT less than 1

This might converge OR diverge.

f(-2)= |$\frac{-2-3}{5}$|

= |-5/5| = 1 is NOT less than 1

This might converge OR diverge.

Oh! THIS is why the points must be manually tested! I did not understand why before!
So now I need to do the integral test?

 

[Infrared]

You have to do another test, but there are other tests easier than the integral test. You will see that one of -2 and 8 will give you something diverges by comparison with harmonic series and the other will converge by alternating series test.

Edit: You don't have to check x=3 because that is safely inside your convergence set. It is just the endpoints that you need to do this for.
Also, you should not use 'f' to denote the ratio as it is already being used for the series itself.

 

[SammyS]

@ Lo.Lee.Ta. ,

For what values of x, is the series an alternating series ?

 

[Lo.Lee.Ta.]

AGH! The integral test seems like too much of a pain here!

Attempt to take integral of 3(x-3)³/[(n+1)(5ⁿ)]:


[(n5ⁿ + 5ⁿ)(3n(x-3)^n-1*1) - (3(x-3)ⁿ)(n²5^n-1 + n5^n-1)]/[(n+1)(5ⁿ)]²


I DO NOT WANT TO HAVE TO FIGURE THIS LONG INTEGRAL OUT! D=

Isn't there some other way that I can figure out if f(-2) and f(8) converge or diverge?!

 

[Infrared]

Yes, read my previous post.

 

[Lo.Lee.Ta.]

Please check to see if my process is right! Thanks!

Okay, let me just start with f(-2).

The expression will look like this when -2 is substituted for x:

$\frac{3(-2-3)^{n}}{(n+1)5^{n}}$

= $\frac{3(-5)^{n}}{(n+1)5^{n}}$

Because of that (-5)ⁿ, we have an alternating series. It will start negative, then +, -, ... etc.
(And- would the -5 be considered the r...?)

The two tests to see if an alternating series converges are:

1. a_n+1≤a_n

2. a_n (stuff without the r) goes to 0 as n approaches infinity.

The first situation is satisfied because a larger denominator results in a smaller fraction.

The second situation is also satisfied because:

a_n = 3/(n+1)5ⁿ

lim = 3/(large denominator) = 0
n→∞

I say, yes, it does converge at f(-2).

 

[Lo.Lee.Ta.]

I am a bit confused as to how to test for f(8).

You say I am supposed to compare to a harmonic series.
How am I supposed to know what would be good to compare it to?

How would I decide if I should compare it to 1/(5ⁿ) or 1/(n+1) or 1/n, etc.?

A harmonic series is in the form 1/n, and we also have a 1/n in our expression. So I guess I'll compare it to that?

3(5)ⁿ/(n+1)5ⁿ ≤ 1/n

But this wouldn't even help us. Our expression has a bigger denominator, and therefore, is smaller than our comparison.
And our comparison diverges.

So what am I supposed to do? D:

 

[SammyS]

3(5)ⁿ/(n+1)5ⁿ ≤ 1/n

$\displaystyle \frac{3\cdot 5^n}{(n+1)5^n}=\frac{3}{n+1}$

Does $\displaystyle \ 3\sum_{n=0}^{\infty}\frac{1}{n+1}\$ converge?

 

[Lo.Lee.Ta.]

No, I think that would still diverge... And that comparison wouldn't help me either because it is greater than the original expression... =(

 

[SammyS]

No, I think that would still diverge... And that comparison wouldn't help me either because it is greater than the original expression... =(

$\displaystyle \sum_{n=0}^{\infty}\frac{1}{n+1}\$ is the same as $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\ .$

Look at it term by term .

 

[Infrared]

No, I think that would still diverge... And that comparison wouldn't help me either because it is greater than the original expression... =(

$\frac{3}{x+1} > \frac{1}{x}$ even if you were starting from the same inital x-value. And, even if it didn't, you did start both series by one, you still could use the limit comparison test.

 

[Lo.Lee.Ta.]

Okay, I see that the a_n is divergent using the limit comparison test.

BUT HOW did you know to make the a_n equal only the 3/(x+1)?

Thanks! :)

 

[Infrared]

I'm sorry, my x should have been an n. Hope that clears up any confusion.