<s>[INST] Proof by induction $\sum_{k=1}^{2^n} \frac1k \ge 1 + \frac{n}2 $ Show that for all $n \in  N  , n \ge 1$ :
$\sum_{k=1}^{2^n}  \frac1k  \ge 1 + \frac{n}2 $
Here is what I got:
I. $A(1)$: $\sum_{k=1}^{2} \frac1k = 1.5 \ge 1.5 \;\;   \implies A(1) \text{ is correct}    $
II. $A(n)$ is correct:  $\sum_{k=1}^{2^n}  \frac1k  \ge 1 + \frac{n}2 $
III. Is $A(n+1)$ correct? $\sum_{k=1}^{2^n+1}  \frac1k  \ge 1 + \frac{n+1}2 $
IV.  Now I need the proof but I don't get the solution for it. Trying for hours - so hope someone can give me an answer with a reasonable explanation. Thank you very much! 
 [/INST] Base case: 
Let $n=1$. Then $\sum_{k=1}^{2^1} \frac{1}{k} \geq 1 + \frac{1}{2}$ by evaluation / inspection.
Induction step (examining the $n+1$ case against the $n$ case): 
$(\sum_{k=1}^{2^{n+1}} \frac{1}{k}) - (\sum_{k=1}^{2^n} \frac{1}{k}) = \sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}$
$(1+\frac{n+1}{2}) - (1+\frac{n}{2}) = \frac{1}{2}$
We only need to prove that $(\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}) \geq \frac{1}{2}$.
Note that there are $2^n$ terms in the lefthand summation, where all values of $\frac{1}{k}$ are $\geq \frac{1}{2^{n+1}}$. 
Therefore $(\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}) \geq (2^n)(\frac{1}{2^{n+1}}) \geq \frac{1}{2}$.
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