Question

$\large\begin{array}{l}\\\\ \textsf{Exerc\'icio relacionado \`a s\'erie harm\^onica e ao fato de esta ser divergente.}\\\\ \textsf{Mostre que}\\\\ \mathsf{\displaystyle\sum_{p=2^{k-1}+1}^{2^k}~\frac{1}{p}~\ge ~\frac{1}{2}}\\\\ \textsf{para todo }\mathsf{k \ge 1,\;k\in\mathbb{N}.}\\ \end{array}$
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Answer 1

 Olá!
 
 Seja $\displaystyle{\mathsf{S_n = \sum_{p = 2^{n - 1} + 1}^{2^n} \frac{1}{p}}}$. 
 
 Parcialmente, temos:

S_1:

$\\ \displaystyle{\mathsf{S_1 = \sum_{p = 2^{1 - 1} + 1}^{2^1} \frac{1}{p}}} \\\\\\ \displaystyle{\mathsf{S_1 = \sum_{p = 2}^{2} \frac{1}{p}}} \\\\\\ \boxed{\mathsf{S_1 = \frac{1}{2}}}$

 Que, satisfaz a desigualdade, pois é igual a 1/2.


S_2:

$\\ \displaystyle{\mathsf{S_2 = \sum_{p = 2^{2 - 1} + 1}^{2^2} \frac{1}{p}}} \\\\\\ \displaystyle{\mathsf{S_2 = \sum_{p = 3}^{4} \frac{1}{p}}} \\\\\\ \boxed{\mathsf{S_2 = \frac{1}{3} + \frac{1}{4}}}$

 Mas, $\mathsf{\frac{1}{3}+\frac{1}{4}\geq\frac{1}{4}+\frac{1}{4}}$. Então,

$\\ \mathsf{\frac{1}{3}+\frac{1}{4}\geq\frac{1}{4}+\frac{1}{4}}\\\\\\ \mathsf{S_2 \geq \frac{1}{4} + \frac{1}{4}} \\\\\\ \mathsf{S_2 \geq \frac{2}{4}} \\\\\\ \boxed{\boxed{\mathsf{S_2 \geq \frac{1}{2}}}}$

S_3:

$\\ \displaystyle{\mathsf{S_3 = \sum_{p = 2^{3 - 1} + 1}^{2^3} \frac{1}{p}}} \\\\\\ \displaystyle{\mathsf{S_3 = \sum_{p = 5}^{8} \frac{1}{p}}} \\\\\\ \boxed{\mathsf{S_3 = \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}}$

Mas, $\mathsf{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\geq\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}$. Então,

$\\ \mathsf{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\geq\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}\\\\\\ \mathsf{S_3\geq\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}\\\\\\ \mathsf{S_3 \geq \frac{4}{8}} \\\\\\ \boxed{\boxed{\mathsf{S_3 \geq \frac{1}{2}}}}$

(...)

S_n:

$\\ \displaystyle{\mathsf{S_n = \sum_{p = 2^{n - 1} + 1}^{2^n} \frac{1}{p}}} \\\\\\ \boxed{\mathsf{S_n = \frac{1}{2^{n - 1} + 1} + \frac{1}{2^{n - 1} + 2} + \frac{1}{2^{n - 1} + 3} + ... + \frac{1}{2^n}}}$

Daí,

$\\ \mathsf{\left(\frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\frac{1}{2^{n-1}+3}+...+\frac{1}{2^n}\right)\geq\left(\frac{1}{2^n}+\frac{1}{2^n}+\frac{1}{2^n}+...+\frac{1}{2^n}\right)}\\\\\\ \mathsf{S_n \geq \frac{2^{n - 1}}{2^n}} \\\\\\ \mathsf{S_n \geq \frac{2^n \cdot 2^{- 1}}{2^n}} \\\\\\ \mathsf{S_n \geq 2^{- 1}} \\\\\\ \boxed{\boxed{\mathsf{S_n \geq \frac{1}{2}}}}$

 Daí, concluímos que, de facto,
 
$\displaystyle{\mathsf{\sum_{p = 2^{k - 1} + 1}^{2^k} \frac{1}{p}\geq\frac{1}{2}}}$