Recent content by d125q

Oh, damn. I must be blind. So, \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 in which case, 2^{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 2 You're awesome, guys.

Indeed. I am still in high school, so we kind of shun the sigma and pi operators, but they surely make the whole deal a lot easier. So, does my above post stand?

If logic serves me well, 1 / 2^n approaches zero, so the whole product equals 1. Is that right (sorry, but I'm rather dizzy right now)? Thanks a million!

The following limit is to be solved: $$ \lim_{n\to\infty}(\sqrt{2} \cdot \sqrt[4]{2} \cdot \sqrt[8]{2} \cdot \sqrt[16]{2} \cdot … \cdot \sqrt[2^n]{2}) $$ all of my attempts to (properly) solve it so far have been futile. And help will be greatly appreciated! EDIT: Perhaps this should go to...