$S_n=\sum_{k=1}^n\frac{1}{k}$. then $S_{2^n}$=? Let $S_n$=$\sum_{k=1}^n\frac{1}{k}$. which of the following is true?


*

*$S_{2^n}\ge\frac{n}{2}$ for every n$\ge1$.

*$S_n$ is a bounded sequence.

*$|S_{2^n}-S_{2^{n-1}}|\to0$ as n$\to\infty$.

*$\frac{S_n}{n}\to1$ as n$\to\infty$.


I have a confusion that whether $S_{2^n}$=$\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n}$? 
or
$S_{2^n}$=$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^n}$? 

The second interpretation is correct: $S_{2^n}=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^n}$. 
Hint for your questions: note that
$$S_{2^{n+1}}=S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+k}\geq S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+2^n}=S_{2^{n}}+\frac{1}{2}.$$
Hence $S_{2^{n+1}}-S_{2^{n}}\geq 1/2$ and 3) is false.
Show by induction that 1) holds and therefore 2) is false. 
What about 4)? 
P.S. 4) is false. We have that
$$S_{2^{n+1}}=S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+k}< S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{k}=2S_{2^{n}}.$$ 
Hence the sequence $(S_{2^{n}}/2^n)_n$ is positive and strictly decreasing.Therefore it converges to a non-negative limit $L$. Since $S_{1}/1=1$ it follows that $L<1$  which contradicts 4).
