Proving the Inequality: Finding the Derivative of (1/2)^m for m=2^n

[puzzek]

Homework Statement

prove the inequality.

Homework Equations

(in the attached file.)

The Attempt at a Solution

The derivative of (1/2)^m where m=2^n is exactly what I need. But I can’t find the sum of (1/2)^m (because as it’s not geometric series, m doesn’t run from 0 to inf).
by the way, am I allowed to make a derivative to this sum (I know I need to prove first that it uniformly converges?

- Another way I tried - to find a Riemann sum that suits this sum (I mean - f(x), Delta-X, and points in the intervals, to find an integral of the sum - but couldn't find a good one.)

Thanks for the help

 

[lippyka]

. Let $S_n = \sum_{m=0}^{2^n} \frac{1}{2^m}$ be the given series. Since each term of the series is positive, by the Comparison Test we have that $S_n \geq \sum_{m=0}^{2^n} \frac{1}{2^{2^n}} = \frac{2^{n+1} - 1}{2^{2^n}}$. Taking the limit as $n \to \infty$, we have $\lim_{n \to \infty} S_n \geq \lim_{n \to \infty} \frac{2^{n+1} - 1}{2^{2^n}} = \frac{1}{2}$. Therefore, $\lim_{n \to \infty} S_n \geq \frac{1}{2}$.