Thread: Sequential Limit Question

1. Sequential Limit Question

I'm not entirely sure what cateogry this question falls under I got it in my Probability class and for some reason none of us can answer it lol. Any help would be greatly appreciated

If b is a positive integer n, such that 1/(2^(n-1)) < b?

2. Originally Posted by AlmasKhan01
I'm not entirely sure what cateogry this question falls under I got it in my Probability class and for some reason none of us can answer it lol. Any help would be greatly appreciated

If b is a positive integer n, such that 1/(2^(n-1)) < b?

Try again: there is no question here.

Tonio

3. Well my Professor was asking questions like can you give a number where this inequality is true. That's the best I can explain it?

Or maybe he was asking If b is a positive integer, give an n such that 1/(2^(n-1)) < b?

He was talking about proving the solution but the question he made us write down the one I wrote I don't understand at all.

Actually he may have asked...... Is it that for any real b there is an integer n such that 1/2^{n-1}

4. Originally Posted by AlmasKhan01
I'm not entirely sure what cateogry this question falls under I got it in my Probability class and for some reason none of us can answer it lol. Any help would be greatly appreciated

If b is a positive integer n, such that 1/(2^(n-1)) < b?
There is no verb in that sentence!

"Given a positive integer b, find an integer n such that $\displaystyle \frac{1}{2^{n-1}}< b$"?

There are no negative numbers here and log is an increasing function so this is pretty much the same as solving the equation.

This inequality is the same as asking for n such that $\displaystyle b< 2^{n-1}$. Since the logarithm is an increasing function, log(b)< log(2^{n-1}= (n-1) log(2). Since log(2) is positive (for positive base) $\displaystyle \frac{log(b)}{log(2)}< n-1$, $\displaystyle n> \frac{log(b)}{log(2)}$.