Paul's Online Notes
Paul's Online Notes
Home / Calculus II / Series & Sequences / Estimating the Value of a Series
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 10.13 : Estimating the Value of a Series

4. Use the Ratio Test and \(n = 8\) to estimate the value of \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{3^{1 + n}}}}{{n\,{2^{3 + 2n}}}}} \).

Show All Steps Hide All Steps

Start Solution

First notice that the terms are positive and so we can use the Ratio Test to do the estimate. Remember that this is a requirement only to use the Ratio Test to get an estimate of the series value and is not an actual requirement to use the Ratio Test to determine if the series converges or diverges.

So, let’s start off with the partial sum using \(n = 8\). This is,

\[{s_8} = \sum\limits_{n = 1}^8 {\frac{{{3^{1 + n}}}}{{n\,{2^{3 + 2n}}}}} = 0.509881435\] Show Step 2

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.