- #1

- 32

- 3

## Homework Statement

Prove that ##\forall n \in \mathbb{N}##

$$\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^n - 1} \leq n \text{ .}$$

## Homework Equations

Peano axioms and field axioms for real numbers.

## The Attempt at a Solution

Okay so my first assumption was that this part in the middle was a partial series of a sequence, that looks something like

$$

1+ \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^n - 1} + \frac{1}{2^{n+1} - 1} \text{,}

$$

so I tried to prove it using this, but I quickly spotted the elephant in the room that the ##\frac{1}{2}## is, since ##\nexists n \in \mathbb{N}## so that ##2^n - 1 = 2##.

Therefore I tried to come up with a reasonable sequence for which such an inequality could hold. Thus far I wrestled with

$$

\sum\limits_{k=1}^{2^n -1} \frac{1}{k} +

\sum\limits_{k=2^n}^{2^{n+1} - 1} \frac{1}{k}\text{,}

$$

but the inequlaity has proven itself to be incredibly difficult to prove via induction.

My first try was to prove the right hand side, therefore to come to a contradiction with

\begin{gather*}

\sum\limits_{k=1}^{2^{n+1} - 1} \frac{1}{k}

\leq

n + \frac{1}{2^n} + \frac{1}{2^n + 1} + \ldots + \frac{1}{2^{n+1} - 1}

> n+1 \text{,}

\end{gather*}

or

$$

\frac{1}{2^n} + \frac{1}{2^n + 1} + \ldots + \frac{1}{2^{n+1} - 1} > 1\text{,}

$$

which is intuitively a contradiciton and I have gone so far as to even write a little program in c that calculated the left side for ##n = 1, 2, \ldots, 10##, where the result was obviously less than 1.

I have no idea how to prove this so far, but I am still trying and will do so until I run out of time and check this thread for hints and update you on the situation when necessary.