Find all real numbers such that the series converges

  • Thread starter tamintl
  • Start date
  • #1
74
0

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
 

Answers and Replies

  • #2
33
0
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?
 
Last edited:
  • #3
74
0
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
33
0
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
74
0
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
33
0
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
74
0
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
33
0
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
74
0
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
33
0
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.
 
Last edited:
  • #11
74
0
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
33
0
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
74
0
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 cant see how I could use them.. Am i missing something here.. There are no values in my original question.
 
  • #14
33
0
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
74
0
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
33
0
Ah sorry I was very tired and mistyped, it should indeed have been -1<y<1.
 
  • #17
74
0
Ah sorry I was very tired and mistyped, it should indeed have been -1<y<1.
great no worries.

Right so ive got it as x>0 and x>-1

hows that?
 
  • #18
33
0
Almost right.
 
  • #19
74
0
Almost right.
Would it just be x>0.. since x≠-1..

So x ε ℝ+ (ie.. positive real numbers)
 
  • #20
33
0
That's exactly right :)
 

Related Threads on Find all real numbers such that the series converges

Replies
12
Views
1K
Replies
1
Views
2K
Replies
3
Views
626
Replies
2
Views
8K
Replies
2
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
21
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
Top