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Section 10.13 : Estimating the Value of a Series

2. Use the Comparison Test and \(n = 20\) to estimate the value of \(\displaystyle \sum\limits_{n = 3}^\infty {\frac{1}{{{n^3}\ln \left( n \right)}}} \).

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Since we are being asked to use the Comparison Test to estimate the value of the series we should first make sure that the Comparison Test can actually be used on this series.

In this case that is easy enough because, for our range of \(n\), the series terms are clearly positive and so we can use the Comparison Test.

Note that it is really important to test these conditions before proceeding with the problem. It doesn’t make any sense to use a test to estimate the value of a series if the test can’t be used on the series. We shouldn’t just assume that because we are being asked to use a test here that the test can actually be used!

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Let’s start off with the partial sum using \(n = 20\). This is,

\[{s_{20}} = \sum\limits_{n = 3}^{20} {\frac{1}{{{n^3}\ln \left( n \right)}}} = 0.057315878\] Show Step 3

Now, let’s see if we can get can get an error estimate on this approximation of the series value. To do that we’ll first need to do the Comparison Test on this series.

That is easy enough for this series once we notice that \(\ln \left( n \right)\) is an increasing function and so \(\ln \left( n \right) \ge \ln \left( 3 \right)\). Therefore, we get,

\[\frac{1}{{{n^3}\ln \left( n \right)}} \le \frac{1}{{{n^3}\ln \left( 3 \right)}} = \frac{1}{{\ln \left( 3 \right)}}\frac{1}{{{n^3}}}\] Show Step 4

We now know, from the discussion in the notes, that an upper bound on the value of the remainder (i.e. the error between the approximation and exact value) is,

\[\begin{align*}{R_{20}} \le {T_{20}} &= \sum\limits_{n = 21}^\infty {\frac{1}{{{n^3}\ln \left( 3 \right)}}} < \int_{{20}}^{\infty }{{\frac{1}{{{x^3}\ln \left( 3 \right)}}\,dx}}\\ & = \mathop {\lim }\limits_{t \to \infty } \int_{{20}}^{t}{{\frac{1}{{{x^3}\ln \left( 3 \right)}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - \frac{1}{{2{x^2}\ln \left( 3 \right)}}} \right)} \right|_{20}^t\\ & = \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{{800\ln \left( 3 \right)}} - \frac{1}{{2{t^2}\ln \left( 3 \right)}}} \right) = \frac{1}{{800\ln \left( 3 \right)}}\end{align*}\] Show Step 5

So, we can estimate that the value of the series is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{s \approx 0.057315878}}\]

and the error on this estimate will be no more than \(\frac{1}{{800\ln \left( 3 \right)}} = 0.001137799\).