# Show if the sequence converges

1. May 15, 2010

### vipertongn

1. The problem statement, all variables and given/known data
showing it is increasing with an upper bound or decreasing
with a lower bound

n/(2^n)

2. Relevant equations

if an >an+1 its underbound
an< an+1 then its upperbound

3. The attempt at a solution
I tried first by finding out the sequence:
a1=1/2
a2=1/2
a3=3/8
a4=1/4
a5=5/32

I'm assuming its decreasing, but I'm not sure if this is monotonic at all. Considering how a1=a2 and then a2>a3 and a3>a4 then a4>a5. I think it's underbound since an>an+1 but the first part threw me off since a1=a2. Someone clarify for me?

2. May 15, 2010

### LCKurtz

It is really trivial to show that sequence is decreasing. Have you tried?

3. May 15, 2010

### vipertongn

Yea I can tell that the sequence is decreasing just by looking at that. I was taught to try (an+1)-an see if that's greater than 0 or not, but its kinda confusing with this equation. then there is also taking the derivative but it also doesn't help much because that ends up being just 1/2^n-1/2^n(log2)

4. May 15, 2010

### LCKurtz

That's good advice. What happens if you work on the inequality an+1 < an with reversible steps? Try putting the 2's on one side by themselves.

5. May 15, 2010

### vipertongn

Ummm... do you think you could give me an example as to how to go about this. I have other problems and this one seems the most simple to actually work with. >.<;; my professor doesn't give good examples and that's usually how I work through my math. Lets see if I can get this right

so n+1/(2n+1< n/2n
That turns into n+1/n < 2n+1/2n If i subtract that...
so...
n+1/n-2n+1/2n<0 so that shows that its decreasing and underbounded (meaning its bounded under something right?

but wait if i do it the other way... isn't 0>2n+1/2n-n+1/n

Last edited: May 15, 2010
6. May 16, 2010

### LCKurtz

You need parentheses around the n+1 on the left don't you? Don't subtract anything. Just look at what you have. Simplify the right side. Compare it to the left.

7. May 17, 2010

### vipertongn

how do i simplify 2n+1/1n? does it become 1 or something?

8. May 17, 2010

### LCKurtz

$$\frac{2^{n+1}}{2^n}\neq\frac{2^{n+1}}{1^n}$$