Chapter 8.3, Problem 39E

a.

To determine

To prove: If $s_n$ is the $n^{th}$ partial sum of the harmonic series, then $s_n\le 1+\ln n$ .

Explanation of Solution

Given Information: A graph of $y=\frac{1}{x}$ must be used to prove the result.

Graph:

  

Proof:

  $f(x)=\frac{1}{x}$is a decreasing function.

Now, $\frac{1}{k}<{\displaystyle \underset{k-1}{\overset{k}{\int }}\frac{1}{x}dx}$

  $\begin{array}{l}\therefore \frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{k}+\mathrm{...}+\frac{1}{n}<{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dx}{x}+}{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dx}{x}+}\mathrm{...}+{\displaystyle \underset{k-1}{\overset{k}{\int }}\frac{dx}{x}+}\mathrm{...}+{\displaystyle \underset{n-1}{\overset{n}{\int }}\frac{dx}{x}}\\ \\ \Rightarrow \frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{k}+\mathrm{...}+\frac{1}{n}<{\displaystyle \underset{1}{\overset{n}{\int }}\frac{dx}{x}}\left(U\sin g{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx+}{\displaystyle \underset{b}{\overset{c}{\int }}f(x)dx={\displaystyle \underset{a}{\overset{c}{\int }}f(x)dx}}\right)\\ \\ \Rightarrow \frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{k}+\mathrm{...}+\frac{1}{n}<{\left[\ln x\right]}_1^n\\ \\ \Rightarrow \frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{k}+\mathrm{...}+\frac{1}{n}<\ln n\\ \\ \Rightarrow 1+\frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{k}+\mathrm{...}+\frac{1}{n}\le 1+\ln n\\ \\ \Rightarrow s_n\le 1+\ln n\left(\because s_n=1+\frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{k}+\mathrm{...}+\frac{1}{n}\right)\end{array}$

Hence proved.

b.

To determine

To prove: The sum of the first million terms of harmonic series is less than $15$ and the sum of the first billion terms is less than $22$ .

Explanation of Solution

Given Information: Result from part (a).

Proof:

From part (a), $s_n\le 1+\ln n$ .

The equality occurs when $n=1$ .

Let $m=1million={10}^6$ and $b=1billion={10}^9$ .

Then,

  $\begin{array}{l}{s}_m<1+\ln ({10}^6)=1+6\ln 10\approx 14.816<15\\ \\ \&s_b<1+\ln ({10}^9)=1+9\ln 10\approx 21.723<22\end{array}$

Hence proved.