Appendix: Accelerating convergence for other series


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 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

  \[ R_ n = T_ n - \frac{1}{8n} + \frac{1}{ 16 n^2} - \frac{3}{128 n^3} + \frac{1}{256 n^4}. \]    

$n$

$T_ n$

$P_ n$

$Q_ n$

$R_ n$

1

0.8666666666666

0.741666666666667

0.804166666666667

0.784635416666667

5

0.8080789523513

0.783078952351398

0.785578952351398

0.785397702351398

10

0.7972961955693

0.784796195569399

0.785421195569399

0.785398148694399

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 of the same series.

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