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Expert-verified Found in: Page 454 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (a) Use a graph of $$y = \frac{1}{x}$$to show that if$${S_n}$$is the $${n^{th}}$$partial sum of the harmonic series, then$${S_n} \le 1 + \ln n$$.(b) The harmonic series diverges, but very slowly. Use part(a) to show that the sum of the first million term is less than$$15$$and the sum of the first billion terms is less than$$22$$.

(a) By using$$y = \frac{1}{x}$$ graph, prove that$${S_n} \le 1 + \ln n$$.

(b) BY using part(a) we proved that the sum of the first million term is less than$$15$$and the sum of the first billion terms is less than$$22$$.

See the step by step solution

## plotting graph using$$y = \frac{1}{x}$$:

We take$$f\left( x \right) = \frac{1}{x}$$so$$\sum\limits_{n = 1}^\infty {\frac{1}{n}}$$is a harmonic series. ## Prove that $${S_n} \le 1 + \ln n$$:

From figure,

$$\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ........... + \frac{1}{n} \le \int_1^n {\frac{1}{x}dx = \ln n}$$.

So,

$${S_n} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ........... + \frac{1}{n} \le 1 + \ln n$$

$${S_n} \le 1 + \ln n$$.

## Show that$${S_{10}}^9 < 22$$and$${S_{10}}^6 < 15$$:

Since$${10^6} = 1million$$ $${10^9} = 1million$$

If number of terms$$n = {10^6}$$then sum$${S_{10}}^6 \le 1 + \ln {10^6} \approx 14.82 < 15$$.

So $${S_{10}}^6 < 15$$.

If number of terms, $$n = {10^9}$$.

The sum$${S_{10}}^9 \le 1 + \ln {10^9} \approx 21.72 < 22$$.

So, $${S_{10}}^9 < 22$$.

Therefore, (a) $${S_n} \le 1 + \ln n$$is proved.

b) We proved sum of the first million terms is less than$$15$$ and sum of the first billion terms is less than$$22$$.

That is, $${S_{10}}^6 < 15$$and$${S_{10}}^9 < 22$$ hence proved. ### Want to see more solutions like these? 