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March 14th, 2016, 07:04 AM  #1 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Acceleration of a sum of a series for Pi Day
. For the series $\displaystyle \ \ \ 4\bigg(1 \  \ \frac{1}{3} \ + \ \frac{1}{5} \  \ \frac{1}{7} \ + \ \ ... \ \bigg), \ \ $ 628 terms are needed to equal to pi correct to two decimal places for the first time for an even number of terms. (An even number of terms always gives a sum that is less than pi.) When 628 terms are used, the sum equals 3.1400002979... Source: WolframAlpha                                                                           I "discovered" a tweak for a related sum that allows for only eight terms to sum to two correct decimal places for pi. $\displaystyle \ \ \ 4\bigg[1 \  \ \frac{1}{3} \ + \ \frac{1}{5} \  \ \frac{1}{7} \ + \ \frac{1}{9} \  \ \frac{1}{11} \ + \ \frac{3}{4}\bigg(\frac{1}{13}\bigg) \  \ \frac{1}{4}\bigg(\frac{1}{15}\bigg) \ \bigg] \ \approx \ \ $3.1401487 The coefficients on the last two terms can be used on many other convergent alternating series to accelerate their sums. 
March 14th, 2016, 09:29 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra 
Here's a different approach Richardson Extrapolation. Let $S = \lim \limits_{n \to \infty} S_n$ where $S_n = 4\sum \limits_{k=1}^n {(1)^{n+1} \over 2n+1}$ Then we assume that for odd $n$ (to get a monotonic sequence) $S_n = S + {a_1 \over n} + {a_2 \over n^2} + \ldots$. Now, if $n$ is very large (as we would like for our estimate of $\pi$) then $\frac1n$ is very small and so we can write $$\begin{aligned}S_n &\approx S + {a_1 \over n} \\ S_{n+2} &\approx S + {a_1 \over n+2} \end{aligned}$$ And now we have two (approximate) equations in two unknowns ($S$ and $a_1$) so we can eliminate $a_1$ to get $$(n+2)S_{n+1}  (n+2)S \approx nS_n  nS \implies S \approx \frac12\big((n+2)S_{n+2}nS_n\big)$$ This is the first Richardson extrapolation for the series. We can improve the approximation by taking the first three terms of the series for $S_n$, $S_{n+2}$ and $S_{n+4}$ and eliminating $a_1$ and $a_2$. This is the second Richardson extrapolation $$S \approx \frac14 \cdot \frac12 \big( (n+4)^2S_{n+4}  2(n+2)^2S_{n+2} + n^2S_n \big)$$ We can then get the following estimates for $\pi$: $$\begin{matrix} n & S_n & \text{1st Richardson} & \text{2nd Richardson} & \text{3rd Richardson} & \text{4th Richardson} \\ 1 & 4.000000 & 3.200000 & 3.136508 & 3.140445 & 3.141557 \\ 3 & 3.466667 & 3.149206 & 3.139882 & 3.141434 & 3.141600 \\ 5 & 3.339683 & 3.143878 & 3.140917 & 3.141555 & 3.141596 \\ 7 & 3.283738 & 3.142562 & 3.141265 & 3.141580 & 3.141594 \end{matrix}$$ 

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acceleration, day, series, sum 
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