Inequality about the most common divergent series

Question

We know that:

$1+\frac{1}{2}+\frac{1}{3}+....$ is a divergent series.

I have a small problem about this series.

Show that: $1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2^{2006}} >50$

How to prove this one?

Can you use a basic knowledge to explain for a 14 years old student?

Answer 1

Notice $1+\frac{1}{2}+\dots + \frac{1}{n} =\int\limits_{i=1}^{n+1} \frac{1}{\lfloor x \rfloor } \geq \int\limits_{i=1}^{n+1} \frac{1}{ x } = \ln(n+1)$

We conclude your sum is at least $\ln(2^ {2006} +1)$ and the rest is easy.

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A basic way to prove it is by noticing

$\frac{1}{2^ n} + \frac{1}{2^n + 1} + \dots + \frac{1}{2^{n+1}-1}> \frac{2^ n}{2^{n+1}} = \frac{1}{2}$ and notice we can split the sum into way more than $100$ of these sums.

Answer 2

You even have more: \begin{align} 1&+\frac 12+\underbrace{\frac 13+\frac 14}_{>\tfrac24=\tfrac12}+\underbrace{\frac 15+\frac 16+\frac17+\frac18}_{>\tfrac48=\tfrac12}+\underbrace{\frac 19+\frac 1{10}+\frac1{11}+\frac1{12}+\frac 1{13}+\frac1{14}+\frac1{15}+\frac1{16}}_{>\tfrac8{16}=\tfrac12}+\\ &+\dotsm+\underbrace{\frac 1{2^{2005}+1}+\frac 1{2^{2005}+2}\dotsm+\frac1{2^{2006}}}_{>\tfrac{2^{2005}}{2^{2006}}=\tfrac12} >1+2006\cdot\frac12=\color{red}{1004}. \end{align}

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