<s>[INST] Prove that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1}I tried using mathematical induction to prove this, but the problem I faced was that there are a lot of numbers between $\frac{1}{2^k-1}$ and $\frac{1}{2^{k+1}-1}$. Is it possible to prove this with induction or is there a better method?
 [/INST] Given the hints above, we have during the induction process that
$$\frac{1}{2^n}+\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^{n+1}-1}<1$$
If we multiply both sides by $2^n$ and note that there are $2^n$ terms being added, we have that
\begin{eqnarray*}
\frac{2^n}{2^n}+\frac{2^n}{2^n+1}+\frac{2^n}{2^n+2}+...+\frac{2^n}{2^{n+1}-1}&<&2^n\\1+\frac{2^n+1-1}{2^n+1}+\frac{2^n+2-2}{2^n+2}+...+\frac{2^n+(2^n-1)-(2^n-1)}{2^{n}+2^n-1}&<&2^n
\\1+1-\frac{1}{2^n+1}+1-\frac{2}{2^n+2}+...+1-\frac{2^n-1}{2^{n+1}-1}&<&2^n\\2^n-\frac{1}{2^n+1}-\frac{2}{2^n+2}-...-\frac{2^n-1}{2^{n+1}-1}&<&2^n\\-\left(\frac{1}{2^n+1}+\frac{2}{2^n+2}+...+\frac{2^n-1}{2^{n+1}-1}\right)&<&0
\end{eqnarray*}
which is true for all $n\ge2$
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