20. Convergence of Positive Series

g. Approximating the Series

4. Improved Comparison Bounds

Assume \(0 \lt a_n \lt b_n\) for \(n \ge n_o\) and \(\displaystyle \sum_{n=n_o}^\infty b_n\) converges. When the series \(\displaystyle S=\sum_{n=n_o}^\infty a_n\) is approximated by the partial sum \(\displaystyle S_k=\sum_{n=n_o}^k a_n\), the series \(\displaystyle \sum_{n=k+1}^\infty b_n\) is an upper bound on the error \(E_k=S-S_k\). However, the approximation to the series can be improved and the error estimate can be reduced if we also use a comparison series which is a lower bound on the error.

In a previous example, we approximated the series \(\displaystyle S=\sum_{n=1}^\infty \dfrac{2+\sin n}{2^n}\) by the \(10^\text{th}\) partial sum \(\displaystyle S_{10}=\sum_{n=1}^{10} \dfrac{2+\sin n}{2^n}\approx 2.5914\) and showed the error \(\displaystyle E_{10}=S-S_{10}=\sum_{n=11}^\infty \dfrac{2+\sin n}{2^n}\) is less than the comparison bound \(\displaystyle \sum_{n=11}^\infty \dfrac{3}{2^n}\approx.00293\). Improve the approximation by using upper and lower comparison bounds on the error \(E_{10}\).

Since \(-1 \le \sin n \le 1\), we have \(1 \le 2+\sin n \le 3\) and \(\dfrac{1}{2^n} \le \dfrac{2+\sin n}{2^n} \le \dfrac{3}{2^n}\). Thus: \[ \sum_{n=11}^\infty \dfrac{1}{2^n} \le E_{10} =\sum_{n=11}^\infty \dfrac{2+\sin n}{2^n} \le \sum_{n=11}^\infty \dfrac{3}{2^n} \] The upper bound is \[ U=\sum_{n=11}^\infty \dfrac{3}{2^n}=\dfrac{\dfrac{3}{2^{11}}}{1- \dfrac{1}{2}}=\dfrac{3}{2^{10}}\approx .00293 \] The lower bound is \[ L=\sum_{n=11}^\infty \dfrac{1}{2^n}=\dfrac{\dfrac{1}{2^{11}}}{1- \dfrac{1}{2}}=\dfrac{1}{2^{10}}\approx .00098 \] Thus \(L \lt E_{10} \lt U\) and \(S_{10}+L \lt S \lt S_{10}+U\). So a better estimate for the sum is the midpoint of the interval \([S_{10}+L,S_{10}+U]\): \[ S\approx\overline{S_{10}}\equiv S_{10}+\dfrac{U+L}{2} \] In this problem \[\begin{aligned} \sum_{n=1}^\infty \dfrac{2+\sin n}{2^n} &\approx \overline{S_{10}} =2.5914+\dfrac{.00293+.00098}{2}=2.5934 \\ \end{aligned}\] The error in this approximation is less than half of the width of the interval \([S_{10}+L,S_{10}+U]\): \[ \overline{E_{10}}\equiv S-\overline{S_k} \lt \dfrac{U-L}{2} \] In this problem \[ \overline{E_{10}} \lt \dfrac{.00293-.00098}{2}\approx.000975 \] Notice, we have reduced the error from \(U=.00293\) to \(\dfrac{U-L}{2}=.000975\).

To summarize (and generalize) these results, we have

Assume \(0 \lt c_n \lt a_n \lt b_n\) 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.

For \(k \gt n_o\), let \(\displaystyle U_k=\sum_{n=k+1}^\infty b_n\) and \(\displaystyle L_k=\sum_{n=k+1}^\infty c_n\).

  1. If the series \(\displaystyle S=\sum_{n=n_o}^\infty a_n\) is approximated by \(\displaystyle S_k=\sum_{n=n_o}^k a_n\), then the error in the approximation, \(\displaystyle E_k=S-S_k=\sum_{n=k+1}^\infty a_n\), is less than \(U_k\).
  2. If the series \(\displaystyle S=\sum_{n=n_o}^\infty a_n\) is approximated by \(\overline{S_k}=S_k+\dfrac{U_k+L_k}{2}\), then the error in the approximation, \(\displaystyle \overline{E_k}=S-\overline{S_k}\), is less than \(\begin{aligned} \displaystyle \dfrac{U_k-L_k}{2} \end{aligned}\).

In a previous exercise, we approximated the series \(\displaystyle S=\sum_{n=1}^\infty \dfrac{1}{2^n+1}\) by the \(10^\text{th}\) partial sum \(\displaystyle S_{10}=\sum_{n=1}^{10} \dfrac{1}{2^n+1}\approx.7635\) and showed the error \(\displaystyle E_{10}=S-S_{10} =\sum_{n=11}^\infty \dfrac{1}{2^n+1}\) is less than the comparison bound \(\displaystyle \sum_{n=11}^\infty \dfrac{1}{2^n}\approx .00098\). Improve the approximation by using upper and lower comparison bounds on the error \(E_{10}\).

The improved approximation is \(\overline{S_{10}}=0.7643\)
and the error is \(\overline{E_{10}} \lt .00024\).

The upper bound on the error \(E_{10}\) is \[ U=\dfrac{1}{2^{10}}\approx.00098 \] To put a lower bound on the error, we note that \[ \dfrac{1}{2^n+1} \gt \dfrac{1}{2^n+2^n}=\dfrac{1}{2^{n+1}} \] So \[\begin{aligned} E_{10}&=\sum_{n=11}^\infty \dfrac{1}{2^n+1} \ge \sum_{n=11}^\infty \dfrac{1}{2^{n+1}} \\ &=\dfrac{\dfrac{1}{2^{12}}}{1-\dfrac{1}{2}} =\dfrac{1}{2^{11}}\approx .00049 \end{aligned}\] Thus, the lower bound on the error \(E_{10}\) is \[ L=\dfrac{1}{2^{11}}\approx.00049 \] Our new approximation is \[\begin{aligned} \overline{S_{10}}&=S_{10}+\dfrac{U+L}{2} \\ &=0.7635+\dfrac{.00098+.00049}{2} =0.7643 \end{aligned}\] and the error in this new approximation is \[ \overline{E_{10}}<\dfrac{U-L}{2}=\dfrac{.00098-.00049}{2}=.00024 \] Notice we have reduced the error from \(U=.00098\) to \(\dfrac{U-L}{2}=.00024\).

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