# Sequential Limit Question

• Oct 14th 2010, 05:53 PM
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?
• Oct 14th 2010, 07:20 PM
tonio
Quote:

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
• Oct 14th 2010, 08:07 PM
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}
• Oct 15th 2010, 02:37 AM
HallsofIvy
Quote:

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