We can easily play the same game with the other series mentioned in this article. For example take the Gregory/Leibnitz series. There is a subtlety with this case as this is an {\em alternating series} which, as you add it up, takes values which oscillate above and below the value of $\pi/4.$ To avoid this we will define the sum $T_n$ by $$ T_n = 1 - \frac{1}{3} + \frac{1}{5} - \ldots - \frac{1}{4n-1} + \frac{1}{4n+1}. $$ This sum is always slightly greater than $\pi/4$. Then $$T_n = T_{n-1} - \frac{1}{4n-1} + \frac{1}{4n+1} = T_{n-1} - \frac{2}{16n^2 -1}.$$ As before we will assume that $T_n \to T$ and that $$T = T_n + \frac{A_0}{n} + \frac{A_1}{n^2} + \frac{A_2}{n^3} + \ldots. $$ Comparing the two expressions for $T_n$ and $T_{n-1}$ we get $$\frac{2}{16 n^2-1} = A_0 \left( \frac{1}{n} - \frac{1}{(n-1)} \right) + A_1 \left( \frac{1}{n^2} - \frac{1}{(n-1)^2} \right) + A_2 \left( \frac{1}{n^3} - \frac{1}{(n-1)^3} \right) + \ldots $$ Again, as before, we can find the values of the terms $A_n$ by expanding out each of these fractions and comparing powers of $1/n^k$ when $n$ is large. This is slightly more complicated than last time an we must use the result that for large values of $n$ $$\frac{2}{16 n^2-1} = \frac{1}{8 n^2} \left( 1 + \frac{1}{16 n^2} + \frac{1}{256 n^4} + \ldots. \right).$$ Doing this we find that $$A_0 = -\frac{1}{8} , A_1 = \frac{1}{16}, A_2 = -\frac{3}{128}, A_3 = \frac{1}{256}$$ so that $$T = T_n - \frac{1}{8n} + \frac{1}{16 n^2} - \frac{3}{128 n^3} + \frac{1}{256 n^4} - \ldots$$ As before we will compare the approximations to the sum given by $T_n$ and by the corrections to $T$ given by $$P_n = T_n - \frac{1}{8n} ,$$ $$Q_n = T_n - \frac{1}{8n} + \frac{1}{16 n^2},$$ and $$R_n = T_n - \frac{1}{8n} + \frac{1}{ 16 n^2} - \frac{3}{128 n^3} + \frac{1}{256 n^4}.$$ \begin{center} \begin{tabular}{|c|c|c|c|c|} \hline $n$ & $T_n$ & $P_n$ & $Q_n$ & $R_n$ \\ \hline 1 & 0.8666666666666 & 0.741666666666667 & 0.804166666666667 & 0.784635416666667 \\ \hline 5 & 0.8080789523513 & 0.783078952351398 & 0.785578952351398 & 0.785397702351398 \\ \hline 10 & 0.7972961955693 & 0.784796195569399 & 0.785421195569399 & 0.785398148694399 \\ \hline \end{tabular} \end{center} Again if we compare these values to the true value of $\pi/4 = 0.785398163397448 \ldots$ then we see excellent agreement, with real economy of effort. In fact $4 \times R_{10} = 3.141592594777596$ which is a far better estimate for $\pi$ than the one that we obtained earlier by adding up 100 terms {\em of the same series}.
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Appendix: Accelerating convergence for other series
Share this pageWe can easily play the same game with the other series mentioned in this article. For example take the Gregory/Leibnitz series. There is a subtlety with this case as this is an {\em alternating series} which, as you add it up, takes values which oscillate above and below the value of $\pi/4.$ To avoid this we will define the sum $T_n$ by $$ T_n = 1 - \frac{1}{3} + \frac{1}{5} - \ldots - \frac{1}{4n-1} + \frac{1}{4n+1}. $$ This sum is always slightly greater than $\pi/4$. Then $$T_n = T_{n-1} - \frac{1}{4n-1} + \frac{1}{4n+1} = T_{n-1} - \frac{2}{16n^2 -1}.$$ As before we will assume that $T_n \to T$ and that $$T = T_n + \frac{A_0}{n} + \frac{A_1}{n^2} + \frac{A_2}{n^3} + \ldots. $$ Comparing the two expressions for $T_n$ and $T_{n-1}$ we get $$\frac{2}{16 n^2-1} = A_0 \left( \frac{1}{n} - \frac{1}{(n-1)} \right) + A_1 \left( \frac{1}{n^2} - \frac{1}{(n-1)^2} \right) + A_2 \left( \frac{1}{n^3} - \frac{1}{(n-1)^3} \right) + \ldots $$ Again, as before, we can find the values of the terms $A_n$ by expanding out each of these fractions and comparing powers of $1/n^k$ when $n$ is large. This is slightly more complicated than last time an we must use the result that for large values of $n$ $$\frac{2}{16 n^2-1} = \frac{1}{8 n^2} \left( 1 + \frac{1}{16 n^2} + \frac{1}{256 n^4} + \ldots. \right).$$ Doing this we find that $$A_0 = -\frac{1}{8} , A_1 = \frac{1}{16}, A_2 = -\frac{3}{128}, A_3 = \frac{1}{256}$$ so that $$T = T_n - \frac{1}{8n} + \frac{1}{16 n^2} - \frac{3}{128 n^3} + \frac{1}{256 n^4} - \ldots$$ As before we will compare the approximations to the sum given by $T_n$ and by the corrections to $T$ given by $$P_n = T_n - \frac{1}{8n} ,$$ $$Q_n = T_n - \frac{1}{8n} + \frac{1}{16 n^2},$$ and $$R_n = T_n - \frac{1}{8n} + \frac{1}{ 16 n^2} - \frac{3}{128 n^3} + \frac{1}{256 n^4}.$$ \begin{center} \begin{tabular}{|c|c|c|c|c|} \hline $n$ & $T_n$ & $P_n$ & $Q_n$ & $R_n$ \\ \hline 1 & 0.8666666666666 & 0.741666666666667 & 0.804166666666667 & 0.784635416666667 \\ \hline 5 & 0.8080789523513 & 0.783078952351398 & 0.785578952351398 & 0.785397702351398 \\ \hline 10 & 0.7972961955693 & 0.784796195569399 & 0.785421195569399 & 0.785398148694399 \\ \hline \end{tabular} \end{center} Again if we compare these values to the true value of $\pi/4 = 0.785398163397448 \ldots$ then we see excellent agreement, with real economy of effort. In fact $4 \times R_{10} = 3.141592594777596$ which is a far better estimate for $\pi$ than the one that we obtained earlier by adding up 100 terms {\em of the same series}.