Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f = \frac{(1/2)^n}{1+(1/2)^n}$ where $n \geq 1 $ I wanted to give an upper bound the function.

So I did

$f = \frac{(1/2)^n}{1+(1/2)^n} \leq \frac{(1/2)^n}{(1/2)^n} = 1$

Which is right, but then I also did

$f = \frac{(1/2)^n}{1+(1/2)^n} \leq \frac{(1/2)^n}{(1)} = (1/2)^n$ and as $n\to \infty$, the function is bounded by $0$ and this makes no sense at all. I have no idea what I am doing wrong in my algebra, but the solution makes no sense ot me, I couldn't interpret the answer at all

share|improve this question
1  
You showed $\lim_{n\to\infty}f=0$, which makes perfect sense. You found a lower bound for all $n$ while the first step you found an upper bound for all $n$. –  Clayton Jan 4 '13 at 3:29
    
Write it as $f(n)\le (1/2)^n$. What you've shown is that the limit of $f$ as $n$ tends to $\infty$ is $0$. Of course, this doesn't mean that $f$ is always less than or equal to $0$. –  David Mitra Jan 4 '13 at 3:29
    
So it is always bounded by 1 for all n. And then as n gets larger you can get another smaller and smaller bound which is going to zero. Being bounded by 1/128 and by 1 simultaneously is no contradiction. –  Fixed Point Jan 4 '13 at 3:30
    
What happens if i replace (1/2) by (-1/2)? What is behavior and bound then? –  sidht Jan 4 '13 at 3:32
    
@Clayton He did not find a lower bound for all n. –  Calvin Lin Jan 4 '13 at 3:33
show 3 more comments

2 Answers

up vote 3 down vote accepted

Why doesn't it make sense? What you have written is correct. In fact, $$0 \leq \dfrac{a^n}{1+a^n} \leq a^n$$ is true for all $a \geq 0$. Hence, if $a < 1$, we have that $$\lim_{n \to \infty} \dfrac{a^n}{1+a^n} = 0$$

EDIT

For $a < 0$, we will split it into three cases. For $a \in (-1,0)$, we have $$\lim_{n \to \infty} \dfrac{a^n}{1+a^n} = \dfrac{\lim_{n \to \infty} a^n}{1+ \lim_{n \to \infty} a^n} = 0$$

For $a \in (-\infty,-1)$, we have $$\dfrac{a^n}{1+a^n} = \dfrac1{1+\left(\dfrac1a \right)^n}$$ Hence, $$\lim_{n \to \infty} \dfrac{a^n}{1+a^n} = \lim_{n \to \infty} \dfrac1{1+\left(\dfrac1a \right)^n} = \dfrac1{1+\lim_{n \to \infty} \left(\dfrac1a \right)^n} = 1$$

For $a=-1$, for even $n$, we have $$\dfrac{(-1)^{2k}}{1+(-1)^{2k}} = \dfrac12$$

For $a=-1$, for odd $n$, it blows up.

For $a=-1+\epsilon$, for odd $n$, we have $$\lim_{\epsilon \to 0^+} \dfrac{(-1+ \epsilon)^{2k+1}}{1+(-1+ \epsilon)^{2k+1}} = -\infty$$

For $a=-1-\epsilon$, for odd $n$, we have $$\lim_{\epsilon \to 0^+} \dfrac{(-1- \epsilon)^{2k+1}}{1+(-1- \epsilon)^{2k+1}} = +\infty$$

share|improve this answer
    
Oh I didn't look at the lower bound. Initally I thought I have a sum of positive numbers less than 0. What is the behaviour if $a < 0$? –  sidht Jan 4 '13 at 3:28
    
@sizz Have updated the behavior for $a < 0$. –  user17762 Jan 4 '13 at 3:40
    
What was the problem of just going straight into $n = 2k+1$? What's the small epsilon for? –  sidht Jan 4 '13 at 3:55
    
@sizz for $n=2k+1$, the expression $\dfrac{a^{2k+1}}{1+a^{2k+1}}$ blows up. What I wanted to convey ws that if you approach from the right of $1$ for an odd $n$, it blows to $-\infty$ while approach from left, it blows to $+\infty$. –  user17762 Jan 4 '13 at 3:59
    
Sorry I just want to ask for the case when $a \in (-1, 0)$ Why does $a^n \to 0$ again? Does it have to do with the geometric series' convergence of $|a| < 1$?? –  sidht Jan 4 '13 at 22:29
show 6 more comments

But you are correct! You made no mistake. Consider: when $n$ is large, $(1/2)^n$ is very close to 0, and $1+(1/2)^n$ is close to 1. Then their quotient is close to 0.

For example, take $n=20$. Then you have $$f(n) = \frac{0.00000095367431640625}{1.00000095367431640625} = 0.00000095367340691241.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.