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?
 A: $$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n-1}=1+\left(\frac{1}{2^1}+\frac{1}{2^2-1}\right)+\left(\frac{1}{2^2}+\cdots+\frac{1}{2^3-1}\right)+\cdots+\left(\frac{1}{2^{n-1}}+\cdots+\frac{1}{2^n-1}\right)<n,
$$
since
$$
\frac{1}{2^{k-1}}+\cdots+\frac{1}{2^k-1}<2^{k-1}\cdot\frac{1}{2^{k-1}}=1.
$$
A: HINT: Try a proof by induction. In your inductive step, subract $n=k$ from $n=k+1$ for this expression:
$$\sum_{r=2^k}^{2^{k+1}-1}{\frac 1r}<1$$
See if you can resolve this.
A: $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1}\approx \int_{2}^{2^n} \frac{1}{k-1}dk$$$$ = ln(2^n-1) = S < ln(2^n) =nln2 =0.693147n<n$$ for $n\geq{2}$
Hence proved
A: 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 $n\ge2$
