Question

(a) Find the partial sum S₁₀of the series\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} \) . Use Exercise 33(a) to estimate the error in using S₁₀as an approximation to the sum of series.

(b) Use exercise 33(b) with n=10to give an improved estimate of the sum.

(c) Find a value of n so that \({{\bf{S}}_{\bf{n}}}\)is within 0.00001 of the sum.

Short answer

(a) The partial sum S₁₀ of the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}} \approx 1.082037} \) and estimate error is \(0.000\overline 3 \)

(b) With n=10 estimate of sum is 1.0823285 with error ≤ 0.00004

(c) n =33 so that \({S_n}\)is within 0.00001 of the sum.

Step-by-step solution

Finding Integral of \(\frac{1}{{{x^4}}}\)

First let \(f(x) = \frac{1}{{{x^4}}}\)

Now we find \(\int\limits_n^\infty {f(x)dx = \int\limits_n^\infty {\frac{1}{{{x^4}}}dx} } \)

\(\begin{aligned} &= \mathop {\lim }\limits_{t \to \infty } {\left( {\frac{{{x^{ - 3}}}}{{ - 3}}} \right)_n}\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{ - 1}}{{3{t^3}}} + \frac{1}{{3{n^3}}}} \right)\end{aligned}\)

\(\begin{aligned} &= 0 + \frac{1}{{3{n^3}}} = \frac{1}{{3{n^3}}}\\ &\Rightarrow \int\limits_n^\infty {f(x)dx = \frac{1}{{3{n^3}}}{\rm{ \_\_\_\_\_(1)}}} \end{aligned}\)

Remainder Estimate for the Integral Test:

Now, \({S_{10}} = \frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + \frac{1}{{{4^4}}} + ..... + \frac{1}{{{{10}^4}}} \approx 1.082037\)

By Remainder Estimate for the Integral Test

\({R_{10}} \le \int\limits_{10}^\infty {f(x)dx} \)

\( \Rightarrow {R_{10}} \le \int\limits_{10}^\infty {f(x)dx = \frac{1}{{3{{(10)}^3}}}} {\rm{ [From (1)]}}\)

\( \Rightarrow {R_{10}} \le \frac{1}{{3000}} \approx 0.000\overline 3 \)

So size of error at most \(0.000\overline 3 \)

Improving the Estimate:

For improving our estimate we use:

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

\({\rm{For }}n = 10\)

\( \Rightarrow {S_{10}} + \int\limits_{11}^\infty {f(x)dx \le S \le {S_{10}}} + \int\limits_{10}^\infty {f(x)dx} \)

\({\rm{Since, }}{{\rm{S}}_{10}} \approx 1.082037\)

\({\rm{And from (1) }}\int\limits_1^\infty {f(x)dx = \frac{1}{{3{{(11)}^3}}} \approx 0.000250} \)

\({\rm{And, }}\int\limits_{10}^\infty {f(x)dx = \frac{1}{{3{{(10)}^3}}} \approx 0.000\overline 3 } \)

Estimating Error:

We have,

\( 1.082037 + 0.000250 \le S \le 1.082037 + 0.000\overline 3 \)

\( \Rightarrow 1.08227 \le S \le 1.082370\)

\( {\rm{So, }}S \approx \frac{{1.082287 + 1.082370}}{2} \Rightarrow S \approx 1.0823285\)

\( {\rm{With error = half of the interval }}[1.082287,1.082370] \approx 0.00004\)

\( \Rightarrow S \approx 1.0823285{\rm{ with error }} \le 0.00004\)

Find n:

We have to find n such that \({R_n}\)≤ 0.00001

Since, \({R_n} \le \int\limits_n^\infty {\frac{1}{{{x^4}}}dx = \frac{1}{{3{n^3}}}} \)

We want \(\frac{1}{{3{n^3}}} < 0.00001\)

\({\rm{Or }}\frac{1}{{{n^3}}} < 0.00003\)

\({\rm{Or }}{n^3} > \frac{1}{{0.00003}} \approx 33333.33 \Rightarrow n > 32.18{\rm{ approx}}{\rm{.}}\)

So, we need n = 33 terms to ensure the accuracy within 0.00003