1. ## Tricky limit

Hi all,

I'm trying toevaluate the following limit;

limit as n --> oo (n-1)*(1/2)^(n-1)

My intuition tells me that it should be zero because (1/2)^(n-1) approaches zero faster than (n-1) approaches infinity...but I'm not sure how to show this rigorously.

Can anyone help?

2. Hi.

Originally Posted by raheel88
Hi all,

I'm trying toevaluate the following limit;

limit as n --> oo (n-1)*(1/2)^(n-1)

My intuition tells me that it should be zero because (1/2)^(n-1) approaches zero faster than (n-1) approaches infinity...but I'm not sure how to show this rigorously.

Can anyone help?
This is correct.

Do you know that $a^x = e^{LN(a)*x}$ ?

I think that's all you need to know, because we have

$(n-1)*(1/2)^{(n-1)} = (n-1)*e^{\displaystyle LN(0.5)*(n-1)} = (n-1)e^{\displaystyle -0.69*(n-1)}$

And we are done with the same argument as above. exp(-x) tends faster to zero than x to infty. (Do you know this result? Are you allowed to use it?)

Rapha

3. Originally Posted by Rapha
Hi.

This is correct.

Do you know that $a^x = e^{LN(a)*x}$ ?

I think that's all you need to know, because we have

$(n-1)*(1/2)^{(n-1)} = (n-1)*e^{\displaystyle LN(0.5)*(n-1)} = (n-1)e^{\displaystyle -0.69*(n-1)}$

And we are done with the same argument as above. exp(-x) tends faster to zero than x to infty. (Do you know this result? Are you allowed to use it?)

Rapha

Hi rapha,

Yes that result rings a bell and I'm definitely allowed to use it. I seem to be a little rusty in analysis .

Thanks for the super-quick response by the way!

raheel88

4. Originally Posted by raheel88
Hi all,

I'm trying toevaluate the following limit;

limit as n --> oo (n-1)*(1/2)^(n-1)

My intuition tells me that it should be zero because (1/2)^(n-1) approaches zero faster than (n-1) approaches infinity...but I'm not sure how to show this rigorously.

Can anyone help?
Yes, it goes to zero quite quickly.

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

$=2\lim_{n\rightarrow\infty}\frac{n}{2^n}-2\lim_{n\rightarrow\infty}\frac{1}{2^n}=0-0$