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Q35E

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Found in: Page 453

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# (a) Use the sum of the first 10 terms and Exercise 33(a) to estimate the sum of the series$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ . How good is this estimate?(b) Improve this estimate using Exercise 33(b) with n = 10(c) Find a value of n that will ensure that the error in the approximation $$S \approx {S_n}$$ is less than 0.01

(a) The sum of series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ is 1.549768 with error at most 0.1

(b) When n = 10, estimate $$S \approx 1.64522$$ with error ≤ 0.005

(c) Value of n > 1000 that will ensure the error approx $$S \approx {S_n}$$ is less than 0.01

See the step by step solution

## (a) Applying Integral on the Given Function:

Let the associated function $$f(x) = \frac{1}{{{x^2}}}$$ is positive and continuous then

$$f(x) = \frac{{ - 2}}{{{x^3}}} < 0{\rm{ for }}x \ge 1$$

Therefore on applying integral it gives an upper bound for the error using the mth partial sum as:

$${R_\infty } = \sum\limits_{n \to 1}^\infty {\frac{1}{{{n^2}}} \le \int\limits_1^\infty {\frac{1}{{{x^2}}}dx} }$$

$${\rm{ }} = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{ - 1}}{x}} \right]_m^b = \frac{1}{m} - \mathop {\lim }\limits_{b \to \infty } \frac{1}{b} = \frac{1}{m}$$

The estimate using the sum of the first ten terms is

$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}} \approx {S_{10}}}$$

$$= 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + .... + \frac{1}{{{{10}^2}}}$$

$$\approx 1.549768$$

And the error of this estimate is $${R_{10}} \le \frac{1}{{10}}$$ , so the error is at most 0.1

## (b) Improving Estimate with n = 10

We have $${S_n} + \int\limits_{n + 1}^\infty {f(x)dx \le S \le {S_n} + \int\limits_n^\infty {f(x)dx} }$$

Now with n = 10, we have $${S_{10}} + \int\limits_{11}^\infty {\frac{1}{{{x^2}}}dx \( = {S_{10}} + \frac{1}{{11}} \le S \le {S_{10}} + \frac{1}{{10}}$$

$$= 1.549768 + 0.090909 \le S \le 1.549768 + 0.1$$

$$= 1.640677 \le S \le 1.6491768$$

So we get S ≈ 1.64522 which is average of 1.640677 and 1.649768, with error ≤ 0.005 (which is rounded up for half 0.1)

## (c) Finding Value of n

We have $${S_n} \le \int\limits_n^\infty {\frac{1}{{{x^2}}}dx{\rm{ which is equal to }}\frac{1}{n}}$$

$${\rm{That is, }}{{\rm{S}}_n} \le \int\limits_n^\infty {\frac{1}{{{x^2}}}dx} = \frac{1}{n}$$

$${\rm{So }}{{\rm{S}}_n} < 0.01{\rm{ if }}\frac{1}{n} < \frac{1}{{1000}}$$

$${\rm{Hence }}n > 1000$$