# Homework Help: Find all real numbers such that the series converges

1. Nov 5, 2011

### tamintl

1. The problem statement, all variables and given/known data

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

Ʃn=1 (1/n) . [(x-1)/(x+1)]n

converges..

2. Relevant equations

3. 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

2. Nov 5, 2011

### Stimpon

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 $$\sum_{n=1}^{\infty}\frac{1}{n}y^{n}$$ 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 $y$ would make the above sum converge?

Last edited: Nov 5, 2011
3. Nov 5, 2011

### tamintl

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. Nov 5, 2011

### Stimpon

Well if you have seen the radius of convergence then you should have seen the formula $$r=lim_{n\rightarrow\infty}|\frac{C_{n}}{C_{n+1}}|$$ so can you see how to use that to find the values of $y$?

And them presuming you can see it, do you see how to find the values of $x$ from that?

5. Nov 5, 2011

### tamintl

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. Nov 5, 2011

### Stimpon

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

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

And $\frac{n+1}{n}$ doesn't diverge.

7. Nov 5, 2011

### tamintl

i don't understand. :/

8. Nov 5, 2011

### Stimpon

Okay so $$\sum_{n=1}^{\infty}\frac{1}{n}y^{n}$$ is of the form $\sum_{n=1}^{\infty}C_{n}y^{n}$ where $C_{n}=\frac{1}{n}$

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

9. Nov 5, 2011

### tamintl

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. Nov 5, 2011

### Stimpon

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

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

Last edited: Nov 5, 2011
11. Nov 5, 2011

### tamintl

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. Nov 5, 2011

### Stimpon

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. Nov 5, 2011

### tamintl

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. Nov 5, 2011

### Stimpon

Well you have that if you want $$\sum_{n=1}^{\infty}\frac{1}{n}y^{n}$$ to converge then you must have $-4<y<4$

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

15. Nov 6, 2011

### tamintl

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. Nov 6, 2011

### Stimpon

Ah sorry I was very tired and mistyped, it should indeed have been -1<y<1.

17. Nov 6, 2011

### tamintl

great no worries.

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

hows that?

18. Nov 6, 2011

### Stimpon

Almost right.

19. Nov 6, 2011

### tamintl

Would it just be x>0.. since x≠-1..

So x ε ℝ+ (ie.. positive real numbers)

20. Nov 6, 2011

### Stimpon

That's exactly right :)