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 $$T_n=1-\frac{1}{3}+\frac{1}{5}-\dots -\frac{1}{4n-1}+\frac{1}{4n+1}.$$ 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 $$T=T_n+\frac{A_0}{n}+\frac{A_1}{n^2}+\frac{A_2}{n^3}+\dots .$$ Comparing the two expressions for $T_n$ and $T_{n-1}$ we get $$\frac{2}{16n^2-1}=A_0(\frac{1}{n}-\frac{1}{(n-1)})+A_1(\frac{1}{n^2}-\frac{1}{(n-1{)}^2})+A_2(\frac{1}{n^3}-\frac{1}{(n-1{)}^3})+\dots$$ 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 $n$ $$\frac{2}{16n^2-1}=\frac{1}{8n^2}(1+\frac{1}{16n^2}+\frac{1}{256n^4}+\dots .).$$ 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 $$T=T_n-\frac{1}{8n}+\frac{1}{16n^2}-\frac{3}{128n^3}+\frac{1}{256n^4}-\dots$$ As before we will compare the approximations to the sum given by $T_n$ and by the corrections to $T$ given by $$P_n=T_n-\frac{1}{8n},$$ $$Q_n=T_n-\frac{1}{8n}+\frac{1}{16n^2},$$ and $$R_n=T_n-\frac{1}{8n}+\frac{1}{16n^2}-\frac{3}{128n^3}+\frac{1}{256n^4}.$$ $$\text{Unknown environment 'center'}$$ Again if we compare these values to the true value of $\pi /4=0.785398163397448\dots$ 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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