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Section 10.13 : Estimating the Value of a Series

1. Use the Integral Test and \(n = 10\) to estimate the value of \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}} \).

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Start Solution

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

First, the series terms are clearly positive so that condition is met.

Now, let’s do a little Calculus I on the following function.

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

The derivative of the function will be negative for \(x > \frac{1}{{\sqrt 3 }} = 0.5774\) and so the function will be decreasing in this range. Because the function and the series terms are the same we can also see that the series terms are decreasing for the range of \(n in our series.

Therefore, the conditions required to use the Integral Test are met! 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!

Show Step 2

Let’s start off with the partial sum using \(n = 10\). This is,

\[{s_{10}} = \sum\limits_{n = 1}^{10} {\frac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}} = 0.392632317\] Show Step 3

Now, to increase the accuracy of the partial sum from the previous step we know we can use each of the following two integrals.

\[\begin{align*}\int_{{10}}^{\infty }{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} & = \mathop {\lim }\limits_{t \to \infty } \int_{{10}}^{t}{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left[ { - \frac{1}{{2\left( {{x^2} + 1} \right)}}} \right]} \right|_{10}^t = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{202}} - \frac{1}{{2\left( {{t^2} + 1} \right)}}} \right] = \frac{1}{{202}}\\ \int_{{11}}^{\infty }{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} & = \mathop {\lim }\limits_{t \to \infty } \int_{{11}}^{t}{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left[ { - \frac{1}{{2\left( {{x^2} + 1} \right)}}} \right]} \right|_{11}^t = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{244}} - \frac{1}{{2\left( {{t^2} + 1} \right)}}} \right] = \frac{1}{{244}}\end{align*}\] Show Step 4

Okay, we know from the notes in this section that if \(s\) represents that actual value of the series that it must be in the following range.

\[\begin{align*}0.392632317 + \frac{1}{{244}} < \, & s < 0.392632317 + \frac{1}{{202}}\\ 0.396730678 < \, & s < 0.397582813 \end{align*}\] Show Step 5

Finally, if we average the two numbers above we can get a better estimate of,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{s \approx 0.397156745}}\]