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 \(\frac{1}{{800\ln \left( 3 \right)}} = 0.001137799\).