Limit of (1-2^n)/(1+2^n): Divergence Explained | Visual Calculus Archive

[jhudson1]

This problem is taken from the visual calculus archives:
http://archives.math.utk.edu/visual.calculus/6/series.15/index.html

Determine whether the series is convergent:

infinity
Ʃ (1-2^n)/(1+2^n)
n=0

Checking this on WolframAlpha I see that the limit of the series is -1, so the series is divergent by the Nth term test. However, I don't understand why the limit is -1.

The explanation given by WolframAlpha seems to rely on some algebraic manipulation which does not immediately jump out at me. Is the process by which WolframAlpha arrived at the answer the most efficient? Should I have seen that? Or is there a better way?

http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+(1-2^n)/(1+2^n)

 

[DryRun]

\lim_{n\to{\infty}} \frac{1-2^n}{1+2^n}=\lim_{n\to{\infty}} \frac{2^n(1/2^n-1)}{2^n(1/2^n+1)} Can you continue from here?

 

[jhudson1]

Sure. Using rules for limits...

\lim_{n\to{\infty}} \frac{2^n(1/2^n-1)}{2^n(1/2^n+1)} = \lim_{n\to{\infty}} \frac{2^n}{2^n}\times\lim_{n\to{\infty}}\frac{(1/2)^n-1}{(1/2)^n+1} \\=\lim_{n\to{\infty}} \frac{2^n}{2^n}\times\left [\lim_{n\to{\infty}}\frac{(1/2)^n}{(1/2)^n} + \lim_{n\to{\infty}}\frac{-1}{+1}\right ]

The first term goes to 0, (inf/inf), which should make the other two terms go to zero regardless of what they come out to be. The third term must be where the final answer is coming from, but why?

 

[HallsofIvy]

"The first term goes to 0"? I'm not sure which you are referring to as the "first term" but both 2^n/2^n and (1/2)^n/(1/2)^n are equal to 1 for all x and so go to 1. Of course, -1/1= -1 for all n so its limit is 1.

 

[jhudson1]

You're right. I was looking at 2^n/2^n and thinking that the result would be \infty/\infty, or undefined or zero. Perhaps it is inappropriate to think of an undefined term as 0?

Either way I was wrong, because it isn't \infty/\infty but rather 2^{\infty}/2^{\infty} which is certainly one.

 

[HallsofIvy]

You're right. I was looking at 2^n/2^n and thinking that the result would be \infty/\infty, or undefined or zero. Perhaps it is inappropriate to think of an undefined term as 0?

Either way I was wrong, because it isn't \infty/\infty but rather 2^{\infty}/2^{\infty} which is certainly one.

Unfortunately, you are still wrong and seem to have a completely wrong idea about limits.
The limit is NOT \infty/\infty nor is it 2^\infty/2^\infty since neither "\infty" nor "2^\infty" since neither of those is a number that can be a limit.

 

[Flumpster]

Of course, -1/1= -1 for all n so its limit is 1.

Why is the limit 1?

 

[jhudson1]

Unfortunately, you are still wrong and seem to have a completely wrong idea about limits.
The limit is NOT \infty/\infty nor is it 2^\infty/2^\infty since neither "\infty" nor "2^\infty" since neither of those is a number that can be a limit.

As n approaches infinity 2^n/2^n becomes larger and larger, but lim_{n\to\infty}2^n/2^n will always be 1 because they grow larger and larger together.

1/1 = 1
32/32=1
4096/4096=1
2^2,000,000/2^2,000,000=1

If you think of it as f(x)=2^x/2^x, no matter what we put in as x (assuming x>0) we will never get anything >1

EDIT: clarify that I am talking about the limit

 

[jhudson1]

Why is the limit 1?

I think he meant -1

 

[Flumpster]

I'm new to this so I'm probably wrong, but here are my two cents:

I'm not sure splitting the limit like that works.

limit (1/2^n-1)/(1/2^n+1) doesn't equal limit (1/2^n)/(1/2^n) + limit (-1/1).

(x+a)/(x+b) doesn't equal x/x + a/b, it equals x/(x+b) + a/(x+b).

(also if your way is correct the limits end up being 1 X [1-1] which doesn't equal -1 which you said you know is the correct answer)

If you factor out the 2^n you're left with

(1/2^n-1)/(1/2^n+1) which if you evaluate as a whole piece and don't split it, gives you -1.

Like I said, probably wrong.

 

[klondike]

dividing both nominator and denominator by 2ⁿ yields \lim _{n\to\infty} \frac{\frac{1}{2^n}-1}{\frac{1}{2^n}+1}, which is trivial to evaluate, isn't it?

This problem is taken from the visual calculus archives:
http://archives.math.utk.edu/visual.calculus/6/series.15/index.html

Determine whether the series is convergent:

infinity
Ʃ (1-2^n)/(1+2^n)
n=0

Checking this on WolframAlpha I see that the limit of the series is -1, so the series is divergent by the Nth term test. However, I don't understand why the limit is -1.

The explanation given by WolframAlpha seems to rely on some algebraic manipulation which does not immediately jump out at me. Is the process by which WolframAlpha arrived at the answer the most efficient? Should I have seen that? Or is there a better way?

http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+(1-2^n)/(1+2^n)

 

[HallsofIvy]

Why is the limit 1?

That was a typo. I mean -1, of course.

 

[jhudson1]

You're right, I can't decompose that fraction that way. Take a look at this:

\lim_{n \to \infty} \frac{2^n}{1+2^n}=\lim_{n \to \infty} \frac{1}{1+2^n}-\lim_{n \to \infty} \frac{2^n}{1+2^n}

\lim_{n \to \infty} \frac{1}{1+2^n}=0

Leaving us with

-\lim_{n \to \infty} \frac{2^n}{1+2^n} = \frac{\lim_{n\to\infty}2^n}{\lim_{n\to\infty}1+2^n} = -1

I may have made a mistake, but I feel good about this answer, but could someone tell me if there is an intermediate step in writing the last two steps? Because:

\lim_{n\to\infty}2^n = \infty\text{ and}\lim_{n\to\infty}1+2^n=\infty
\text{but} \frac{\infty}{\infty}\neq 1

See what I mean?

 

[jhudson1]

dividing both nominator and denominator by 2ⁿ yields \lim _{n\to\infty} \frac{\frac{1}{2^n}-1}{\frac{1}{2^n}+1}, which is trivial to evaluate, isn't it?

Yes. That is very easy and very intuitive, at least in my mind.

Thanks, by the way!