<s>[INST] Proving an inequality using induction Use induction to prove the following:
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^n}\geq1+\frac{n}{2}$
What would the base case be? Would it still be $n=0$
so $\frac{1}{1}+\frac{1}{2}\geq 1+\frac{0}{2}$, which holds true.
then how would you prove for $n$ and $n+1$ to prove the proof with induction?
 [/INST] Yes, the base case here would be $n = 0$, and your assessment is correct, although there is no $\frac{1}{2}$ term in the $n = 0$ case.  
For our inductive step (proving $n+1$ from $n$), consider that we know that 
$$\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2^n} \ge 1 + \frac{n}{2}$$ and are trying to show:
$$\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2^{n+1}} \ge 1 + \frac{n+1}{2}.$$
We thus, subtracting, want to show that:
$$
\frac{1}{2^{n}+1} + \cdots + \frac{1}{2^{n+1}} \ge \frac{1}{2}
$$
Note that there are $2^{n+1} - (2^n + 1) + 1 = 2^n$ terms here, each of which is greater than or equal to $\frac{1}{2^{n+1}}$.  Therefore we get that:
$$
\frac{1}{2^{n}+1} + \cdots + \frac{1}{2^{n+1}} \ge 2^n\left(\frac{1}{2^{n+1}}\right) = \frac{1}{2}.
$$
So we are done.  
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