$ S_n=\sum_{k=1}^n\frac{1}{k}$ then is $S_n$ bounded?

Question

Let , $\displaystyle S_n=\sum_{k=1}^n\frac{1}{k}$. Which of the following is TRUE ?

(A) $\displaystyle S_{2^n}\ge \frac{n}{2}$ for every $n\ge 1$.

(B) $S_n$ is bounded sequence.

(C) $\displaystyle|S_{2^n}-S_{2^{n-1}}|\to 0$ as $n\to \infty$.

(D) $\displaystyle\frac{S_n}{n}\to 1$ as $n\to \infty$.

As the series is divergent , so (B) is FALSE.

If (C) is TRUE then $\{S_n\}$ is a Cauchy sequence, which is NOT possible.

Again , from Cauchy's first limit theorem , $\displaystyle\frac{S_n}{n}\to 0$ as $n\to \infty$. So it is FALSE.

So finally , (A) is correct. Am I correct ?

Answer 1

Yes, $$\sum_{k=1}^{2^n}\frac1k>\int_{1}^{2^n+1}\frac1x\,dx=\ln(2^n+1)>n\ln2>\frac{n}2$$ Actually you can prove the stronger inequality $$\sum_{k=1}^{2^n}\frac1k\ge 1+\frac{n}{2}$$

Answer 2

Yes, A is true. It is true for $n=0$ and at each step $n\to n+1$, each of the additional $2^n$ summands is $\ge \frac1{2^{n+1}}$.

Of course, after this immeditaley $A\to \neg B$, $A\to \neg C$, $A\to \neg D$.

Answer 3

To sum up earlier answers and solution proposed in the question, and to fill in gaps, for the sake of future readers:

$S_n$ is the partial sum of the series $\sum_{k=1}^\infty \frac1k$ which is divergent. Also, $S_n$ is an increasing sequence. So it cannot be bouneded or Cauchy.

Cauchy's first limit theorem: If $\{x_n\}\to k$ as $n\to\infty$, then $\{y_n\}\to k$ as $n\to\infty$, where $y_n=\frac{x_1+\cdots+x_n}{n}$. In this question, take $x_n=\frac1n$, then $y_n={S_n\over n}$.

The inequality used in above answers is: for a decreasing function $f$ $$\int_{x=a}^{b+1}f(x)dx\le\sum_{k=a}^b f(k)\le\int_{x=a-1}^{b}f(x)dx$$ and let $f(k)=\frac1k$, a decreasing function, and let $a=1,b=2^n$