# Simple limit?

1. Mar 1, 2007

### hartigan83

simple limit?!?!?

1. The problem statement, all variables and given/known data
I am trying to prove if a series is convergent.
the series is a(sub n) = (2/3)[2^n - 2^(-n)] from n=1 to infinity.

2. Relevant equations
Ratio test:
the limit as n approached infinity of the absolute value of a(sub n+1) /a(sub n) equals r.
If r is less than one the series converges.
if r is greater than 1 the series diverges.

3. The attempt at a solution

lim as n approaches infinity of the absolute value of
(2/3 * (2^(n+1) - 2 ^(-n+1)))
(2/3 * (2^n - 2^(-n)))
eqauls r
the 2/3 cancel and I get
lim as n approaches infinity of the absolute value of
(2^(n+1) - 2 ^(-n+1))
(2^n - 2^(-n))
eqauls r
this is where I get stuck. Looking at the graph of this function, I can see that the limit is 2. But I dont know how to show it. It has been a long time since I took Calc....
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 1, 2007

### gammamcc

Are you talking about the series \sum a_n or a power series
\sum a_n x^n

or something else
?

3. Mar 1, 2007

### hartigan83

The series sum, sigma notation sorry I just dindt know how to put that on here

4. Mar 2, 2007

### HallsofIvy

Staff Emeritus
Isn't it obvious that the sequence (2/3)(2n- 2-n) does not converge to 0?

5. Mar 2, 2007

### Schrodinger's Dog

a_n = (2/3)[2^n - 2^(-n)] from n=1 to infinity.=

$$a_n=\sum_{n=1}^\infty\frac{2}{3}(2^n - 2^{-n}) \lim_{1\rightarrow\infty}$$

Am I being dim here, but isn't it obvious that the total becomes arbitrarily large as n increases. Therefore it does not converge?

2^n increases as n increases

2^-n decreases as n increases

thus 2/3(2^n-2^-n) becomes increasingly large, thus the sum does not converge, the 2/3 outside the brackets makes no difference to this in this case.

Last edited: Mar 2, 2007
6. Mar 2, 2007

### hartigan83

I thought that the sequence diverged but I have to prove it not just say that it does and I dont know exactally how to show it
Thats why I was using the ration test....but I couldnt figure out how to do the limits.

7. Mar 2, 2007

### hartigan83

right I see that but I am not sure how to "prove it" I guess I need to show that the limit is infinity but I dont know how.

8. Mar 2, 2007

### hartigan83

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

may be it will help if I included the whole problem, maybe I am not doing it correctly at all.

Consider the sequence 1,5/2,21/4,85/8,... defined by s(sub1) =1 , a(sub2)=5/2 and a(subn)= (5/2)a(subn-1) - a(subn-2)

A.) Show that a(subn) = (2/3)[2^n -2^(-n)] for any n greaterthan or equal to 3.

B.) Is the series (sigma notation) from n=1 to infinity a(subn) convergent? (prove your answer)

What I have done:
For A.) I have proven the base case and am working on the inductive step. I know that for the inductive step, I need to show that assuming a(subn) and everything below it is true, that a(subn+1)= (5/2)a(subn) - a(subn-1) = (2/3)[2^(n+1) -2^(-n+1)] , but my algebra isnt working out. so I am still pluggin away at that.:yuck:
For B.) I was thinking that if I show (2/3)[2^n -2^(-n) is divergent, which I can clearly see, but dont know how to prove (I was thinking of using the ratio test as quoted above)then I can conclude that Is the series (sigma notation) from n=1 to infinity a(subn) is divergent.
so can someone help me with this equation if my reasoning is correct::uhh:

lim as n approaches infinity of the absolute value of
(2^(n+1) - 2 ^(-n+1))
(2^n - 2^(-n))
I need help showing my steps for getting the limit.
when I plugged this equation into my calculator, I got a straight line through y=2 so I am thinking this means the limit as n approaches infinity the limit is 2, and thus by the ratio test it is divergent......

9. Mar 2, 2007

### cristo

Staff Emeritus
So, you have $$\lim_{n\rightarrow\infty}\left|\frac{2^{n+1}-2^{1-n}}{2^n-2^{-n}}\right|$$, supposing you have used the ratio test correctly. Now, can you evaluate this limit? Hint: divided top and bottom by 2n

10. Mar 2, 2007

### HallsofIvy

Staff Emeritus
There is a very nice theorem that says if
$$\Sigma_{n=0}^\infty a_n$$
converges then {an} must converge to 0. It's contrapositive is that if {an} does not converge to 0, then $$\Sum_{n=0}^\infty a_n$$ (2/3)(2n+ 2-n) does not converge to 0. That should be easy. (In fact, it goes to infinity!)

11. Mar 3, 2007

### hartigan83

I am not sure how do do it with out getting common denomionators and such... so here is my algebra... I know that all of them have absolute value and the lim next to them but I cant figure out how to use them so I will just write the frations.

2^(2n+2) -1
2^(n+1)
divided by:
2^(2n)-1
2^n

which is
(2^(2n+2)-1)(2^n)
(2^(n+1))(2^(2n)-1)

which is
(2^(2n+2)-1)(2^n)
(2^(2n+1)-2)(2^n)

then the 2^n cancel and I am left with
(2^(2n+2)-1)
(2^(2n+1)-2)

which is
(2^(2n)*(2^2))-1
(2^(2n)*(2))-2

which is
1/2??? so the limit is 1/2?

Last edited: Mar 3, 2007
12. Mar 4, 2007

### tim_lou

as HallsofIvy says, it is quite obvious (painfully) that the series diverge.

consider the series, 1+1+1+1+1+1+1+1+1+1+1.... obviously, it diverges, now consider the original series, every term eventually becomes bigger than 1... so...

Last edited: Mar 4, 2007
13. Mar 4, 2007

### cristo

Staff Emeritus
I'm not sure what you've done there. I said divide top and bottom by 2n to obtain $$\left|\frac{2^{n+1}-2^{1-n}}{2^n-2^{-n}}\right|=\left|\frac{2-2^{1-2n}}{1-2^{2n}}\right|\rightarrow ?$$

If you can see what this tends to, then you can work out your answer using the ratio test. However, Halls has given you a nice theorem which is quicker to use!

14. Mar 4, 2007

### hartigan83

Thanks all I figured it out :)