Find all real numbers such that the series converges

In summary, the original series converges for all real numbers x that satisfy the inequality -1<y<1, where y=(x-1)/(x+1). This can be found by first finding the radius of convergence, r, using the formula r=lim_{n\rightarrow\infty}\frac{C_{n}}{C_{n+1}}, where C_{n}=1/n. Solving for r=1, we get -1<y<1, which can then be used to find the range of values x can take for the series to converge.
  • #1
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Homework Statement



Find all real numbers x ≠ −1 such that the series

Ʃn=1 (1/n) . [(x-1)/(x+1)]n

converges..


Homework Equations





The Attempt at a Solution



I really do not know where to start here. I assume you could first prove it converges sing a test but I'm really not sure.

Regards
 
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  • #2
tamintl said:
I really do not know where to start here. I assume you could first prove it converges sing a test but I'm really not sure.


Well no, you're trying to find x such that the series converges, not prove that the series does converge, which it doesn't for many values of x.

If I asked you to find the radius of convergence for [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] would you know what I was on about? Or if you haven't done radius of convergence yet, would you be perhaps able to see which values of [itex]y[/itex] would make the above sum converge?
 
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  • #3
Stimpon said:
Well no, you're trying to find x such that the series converges, not prove that the series does converge, which it doesn't for many values of x.

If I asked you to find the radius of convergence for [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] would you know what I was on about? Or if you haven't done radius of convergence yet, would you be perhaps able to see which values of [itex]y[/itex] would make the above sum converge?

Yes, i have brief experience with the radius of convergence, however I thought that more often than not you had to use the ratio test with the radius of convergence? In that example would the Radius = ∞?

What values of y would we need for it to converge?

Regards
 
  • #4
Well if you have seen the radius of convergence then you should have seen the formula [tex]r=lim_{n\rightarrow\infty}|\frac{C_{n}}{C_{n+1}}|[/tex] so can you see how to use that to find the values of [itex]y[/itex]?

And them presuming you can see it, do you see how to find the values of [itex]x[/itex] from that?
 
  • #5
Stimpon said:
Well if you have seen the radius of convergence then you should have seen the formula [tex]r=lim_{n\rightarrow\infty}|\frac{C_{n}}{C_{n+1}}|[/tex] so can you use that to find the values of [itex]y[/itex]?
Okay, using the formula you gave me:

lim(n→∞) (n+1)/n . yn/yn+1

= lim(n→∞) (n+1)/n . 1/y

Now what?.. I can see the (n+1)/n diverges..
 
  • #6
tamintl said:
Okay, using the formula you gave me:

lim(n→∞) (n+1)/n . yn/yn+1

= lim(n→∞) (n+1)/n . 1/y

Now what?.. I can see the (n+1)/n diverges..

No sorry I should have stated what the various things were.

If you have something of the form [tex]\sum_{n=1}^{\infty}C_{n}y^{n}[/tex] then the radius of convergence, [itex]r[/itex] is defined by the formula I gave in the last post.

And [itex]\frac{n+1}{n}[/itex] doesn't diverge.
 
  • #7
Stimpon said:
No sorry I should have stated what the various things were.

If you have something of the form [tex]\sum_{n=1}^{\infty}C_{n}y^{n}[/tex] then the radius of convergence, [itex]r[/itex] is defined by the formula I gave in the last post.

And [itex]\frac{n+1}{n}[/itex] doesn't diverge.

i don't understand. :/
 
  • #8
Okay so [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] is of the form [itex]\sum_{n=1}^{\infty}C_{n}y^{n}[/itex] where [itex]C_{n}=\frac{1}{n}[/itex]

Do you see now how to find the radius of convergence for the series?
 
  • #9
Stimpon said:
Okay so [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] is of the form [itex]\sum_{n=1}^{\infty}C_{n}y^{n}[/itex] where [itex]C_{n}=\frac{1}{n}[/itex]

Do you see now how to find the radius of convergence for the series?

Okay.. so using that formula we get as n→∞ lim [ (n+1)/1 . 1/y ]

but since (n+1)/1 converges to 1 then we have... 1/y < 1.. therefore y>1

How's that
 
  • #10
tamintl said:
Okay.. so using that formula we get as n→∞ lim [ (n+1)/1 . 1/y ]

but since (n+1)/1 converges to 1 then we have... 1/y < 1.. therefore y>1

How's that

No, you're not finding the limit of anything involving [itex]y[/itex], you're just finding [tex]r=\lim_{n\rightarrow\infty}\frac{C_{n}}{C_{n+1}}[/tex].

And then from there you'll have [itex]-r<y<r[/itex] as values for which the series converges which you can use to find [itex]x[/itex] in your original question.
 
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  • #11
Stimpon said:
No, you're not finding the limit of anything involving [itex]y[/itex], you're just finding [tex]r=lim_{n\rightarrow\infty}\frac{C_{n}}{C_{n+1}}[/tex].

And then from there you'll have [itex]-r<y<r[/itex] as values for which the series converges which you can use to find [itex]x[/itex] in your original question.

Wow, I'm not functioning today..

So Cn=1/n

So, r = lim (1/n) / (1/n+1)

r = lim (n+1)/n

r=1
...

therefore, -1<y<1
 
  • #12
Yes! Now that you know the range of values y=(x-1)/(x+1) can take, you should be able to immediately see how to find the range of values x can take :)
 
  • #13
Stimpon said:
Yes! Now that you know the range of values y=(x-1)/(x+1) can take, you should be able to immediately see how to find the range of values x can take :)

Thanks for your help.. I still can't see how I could use them.. Am i missing something here.. There are no values in my original question.
 
  • #14
Well you have that if you want [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] to converge then you must have [itex]-4<y<4[/itex]

Then you can simply say [itex]y=\frac{x-1}{x+1}[/itex] and so [itex]-4<\frac{x-1}{x+1}<4[/itex] and from there find the range of values [itex]x[/itex] can take for the original sum in your opening post to converge.
 
  • #15
Stimpon said:
Well you have that if you want [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] to converge then you must have [itex]-4<y<4[/itex]

Then you can simply say [itex]y=\frac{x-1}{x+1}[/itex] and so [itex]-4<\frac{x-1}{x+1}<4[/itex] and from there find the range of values [itex]x[/itex] can take for the original sum in your opening post to converge.

Great. But why have you put the values for r = 4 ?? I thought we agreed that the r=1

Therefore we would have -1<y<1 ?

Thanks for everything
 
  • #16
Ah sorry I was very tired and mistyped, it should indeed have been -1<y<1.
 
  • #17
Stimpon said:
Ah sorry I was very tired and mistyped, it should indeed have been -1<y<1.

great no worries.

Right so I've got it as x>0 and x>-1

hows that?
 
  • #18
Almost right.
 
  • #19
Stimpon said:
Almost right.

Would it just be x>0.. since x≠-1..

So x ε ℝ+ (ie.. positive real numbers)
 
  • #20
That's exactly right :)
 

1. What is a series?

A series is a mathematical sequence in which the terms are added together in a specific order.

2. What does it mean for a series to converge?

A series converges if the sum of its terms approaches a finite value as the number of terms increases. In other words, the series has a finite sum.

3. How can I determine if a series converges or diverges?

There are several tests that can be used to determine the convergence or divergence of a series, such as the ratio test, the root test, and the comparison test. These tests involve analyzing the behavior of the terms in the series and their relationship to each other.

4. Are there specific types of series that always converge?

Yes, there are certain types of series that are guaranteed to converge, such as geometric series, telescoping series, and p-series. However, it is important to note that these series must meet certain criteria in order to converge, and not all series of these types will necessarily converge.

5. Why is it important to find all real numbers such that the series converges?

Determining the values of real numbers for which a series converges can help in solving various mathematical problems and applications, such as in calculating probabilities, approximating values, and analyzing the behavior of functions. It also allows for a deeper understanding of the properties and patterns of different types of series.

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