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Q37E

Expert-verifiedFound in: Page 454

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**\(22\).

We take\(f\left( x \right) = \frac{1}{x}\)so\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \)is a harmonic series.

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,

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

\({S_n} \le 1 + \ln n\).

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.

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