Now, to get an upper bound on the value of the remainder (i.e. the error between the approximation and exact value) we need the following ratio,

\[{r_n} = \frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{3^{2 + n}}}}{{\left( {n + 1} \right)\,{2^{5 + 2n}}}}\frac{{n\,{2^{3 + 2n}}}}{{{3^{1 + n}}}} = \frac{{3n}}{{4\left( {n + 1} \right)}}\]

We’ll also potentially need the limit,

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{3n}}{{4\left( {n + 1} \right)}} = \frac{3}{4}\] Show Step 3

Next, we need to know if the \({r_n}\) form an increasing or decreasing sequence. A quick application of Calculus I will answer this.

\[f\left( x \right) = \frac{{3x}}{{4\left( {x + 1} \right)}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}f'\left( x \right) = \frac{3}{{4{{\left( {x + 1} \right)}^2}}} > 0\]

As noted above the derivative is always positive and so the function, and hence the \({r_n}\) are increasing.

Show Step 4

The upper bound on the remainder is then,

\[{R_8} \le \frac{{{a_9}}}{{1 - L}} = \frac{{\frac{{6561}}{{2,097,152}}}}{{1 - \frac{3}{4}}} = 0.012514114\] Show Step 5

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{s \approx 0.509881435}}\]

and the error on this estimate will be no more than 0.012514114.