20. Convergence of Positive Series

g. Approximating the Series

We now turn to approximating series which can be shown to converge using the Comparison Test.

3. Comparison Bounds

Show that the series \(\displaystyle \sum_{n=1}^\infty \dfrac{2+\sin n}{2^n}\) converges. Sum the first \(10\) terms to approximate the sum of the series. Find an upper bound on the error in the approximation.

Since \(\sin n \le 1\), we have \(2+\sin n \le 3\) and \(\dfrac{2+\sin n}{2^n} \le \dfrac{3}{2^n}\). Thus \(\displaystyle \sum_{n=1}^\infty \dfrac{2+\sin n}{2^n} \le \sum_{n=1}^\infty \dfrac{3}{2^n}\). Since \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{2^n}\) is a geometric series with ratio \(\dfrac{1}{2}\) it converges. By the Simple Comparison Test \(\displaystyle \sum_{n=1}^\infty \dfrac{2+\sin n}{2^n}\) also converges.

We approximate the sum of the series by the \(10^\text{th}\) partial sum \[ \sum_{n=1}^\infty \dfrac{2+\sin n}{2^n}\approx S_{10}= \sum_{n=1}^{10} \dfrac{2+\sin n}{2^n}\approx 2.5914 \] The error in the approximation is \[ E_{10}=\sum_{n=1}^\infty \dfrac{2+\sin n}{2^n} -\sum_{n=1}^{10} \dfrac{2+\sin n}{2^n} =\sum_{n=11}^\infty \dfrac{2+\sin n}{2^n} \] Since the terms \(\dfrac{2+\sin n}{2^n}\) are positive, \(E_{10} \gt 0\). To put an upper bound on the error, we again use the comparison \(\dfrac{2+\sin n}{2^n} \le \dfrac{3}{2^n}\). Thus \[ E_{10}=\sum_{n=11}^\infty \dfrac{2+\sin n}{2^n} \lt \sum_{n=11}^\infty \dfrac{3}{2^n}=\dfrac{\dfrac{3}{2^{11}}}{1- \dfrac{1}{2}}=\dfrac{3}{2^{10}}\approx.00293 \] Therefore, the error in the approximation \(\displaystyle \sum_{n=1}^\infty \dfrac{2+\sin n}{2^n} \approx S_{10}=\sum_{n=1}^{10} \dfrac{2+\sin n}{2^n} \approx 2.5914\) is \(E_{10} \lt .00293\).

To summarize (and generalize) this result, we have

Assume \(0 \lt a_n \lt b_n\) for \(n \ge n_o\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) converges. In other words, \(\displaystyle S=\sum_{n=n_o}^\infty a_n\) converges by the Simple Comparison Test.

If the series \(\displaystyle S=\sum_{n=n_o}^\infty a_n\) is approximated by \(\displaystyle S_k=\sum_{n=n_o}^k a_n\) for some \(k \gt n_o\), then the error in the approximation \(\displaystyle E_k=S-S_k=\sum_{n=k+1}^\infty a_n\) is less than \(\displaystyle \sum_{n=k+1}^\infty b_n\) which is called the comparison bound on the error in the approximation.

Consider the series \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n+1}\).

  1. Show that the series converges.

    Since \(\dfrac{1}{2^n+1} \lt \dfrac{1}{2^n}\), we have \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n+1} \lt \sum_{n=1}^\infty \dfrac{1}{2^n}\). Since \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n}\) is a geometric series with ratio \(\dfrac{1}{2}\) it converges. By the Simple Comparison Test, \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n+1}\) also converges.

  2. Approximate the sum of the series by summing \(10\) terms. Find a comparison bound on the error in this approximation.

    \(S_{10}\approx 0.7635\) and \(E_{10} \lt \dfrac{1}{2^{10}}\approx.00098\)

    The \(10^\text{th}\) partial sum is \[ S_{10}=\sum_{n=1}^\infty \dfrac{1}{2^n+1}\approx \sum_{n=1}^{10} \dfrac{1}{2^n+1}\approx 0.7635 \] (This should probably be done using some type of calculator.) The error in this approximation is \[ E_{10}=\sum_{n=11}^\infty \dfrac{1}{2^n+1} \] To put an upper bound on this error, we again use the comparison \(\dfrac{1}{2^n+1} \lt \dfrac{1}{2^n}\). Thus \[\begin{aligned} E_{10}&=\sum_{n=11}^\infty \dfrac{1}{2^n+1} \lt \sum_{n=11}^\infty \dfrac{1}{2^n} =\dfrac{\dfrac{1}{2^{11}}}{1-\dfrac{1}{2}} \\ &=\dfrac{1}{2^{10}}\approx.00098 \end{aligned}\]

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