$S_n=\sum_{k=1}^n\frac{1}{k}$. then $S_{2^n}$=?

Question

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 $S_{2^n}$=$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^n}$?

Your second interpretation is correct. Notice that $S_{2^n} - S_{2^{n-1}} = a_{2^n} \rightarrow 0$ as $n \rightarrow \infty$ — Hendrra, May 26 '17 at 11:47
Obviously $S_{2^n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n}$ — maverick, May 26 '17 at 11:48
I don't think this question should be closed as off topic, the OP is active in the question (it could be closed as a duplicate but that's a whole another story). — kingW3, May 26 '17 at 12:30
Only 1 is correct according to me. 2 and 3 could be easily rejected by the divergence nature of harmonic series, while 4 is trivially wrong by using the formula $H_n =\ln n +O(1)$. — BAI, May 26 '17 at 12:59

Robert Z · Accepted Answer · 2017-05-26T12:26:01.467

4

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).

edited May 26 '17 at 12:26

answered May 26 '17 at 11:51

Robert Z

145,942

Now i understood the question properly. there are more than one term between $s_{2^n}$ and $S_{2^{n-1}}$. 4) is not true – Priyanka May 26 '17 at 12:08
Yes, 4) is false. Have you a proof? – Robert Z May 26 '17 at 12:09
here $a_n$=$\frac{1}{2^n}\to0$ as n$\to\infty$ then by cauchy's first theorem on limits $\frac{S_n}{n}\to0$ as n$\to\infty$. m i right? – Priyanka May 26 '17 at 12:12
OK. This is a way to show it. – Robert Z May 26 '17 at 12:15
@Priyanka I gave another proof that 4) is false (see my PS). Any more doubts? – Robert Z May 26 '17 at 12:28

score 1 · Answer 2 · answered May 26 '17 at 12:50

$3)$ and $4)$ can be rejected by using the formula $H_n=\ln{n}+\gamma+O(n^{-1})$. That is: $$3) \lim_\limits{n\to\infty}\left[H_{2^n}-H_{2^{n-1}}\right]=\ln2$$ $$4)\lim_\limits{n\to\infty} \frac{H_n}{n}=\lim_\limits{n\to\infty} \frac{\ln{n}+\gamma+O(n^{-1})}{n}=0$$

$S_n=\sum_{k=1}^n\frac{1}{k}$. then $S_{2^n}$=?

2 Answers2