There is an averaging formula, one of many:
\(\displaystyle \tfrac{1}{16}[16a + 15b + 10c + 3d]\), where the a, b, c, d are the consecutive terms of a slowly converging
alternating series.
\(\displaystyle \pi/4 \approx \dfrac{1}{16}\bigg[16(1) + 15(-1/3) + 10(1/5) + 3(-1/7)\bigg]\)
\(\displaystyle \pi \approx 3.1428...\)
You can apply this to 1 - 1/2 + 1/3 - 1/4 to estimate ln(2), or log(2), if you prefer writing it that way, to give it
to two correct rounded decimal places.
Examples of others that it may be used on:
\(\displaystyle 1/1^2 - 1/2^2 + 1/3^2 - 1/4^2\)
\(\displaystyle \dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{4}} \)
\(\displaystyle \tfrac{1}{16}[16a + 15b + 10c + 3d]\), where the a, b, c, d are the consecutive terms of a slowly converging
alternating series.
\(\displaystyle \pi/4 \approx \dfrac{1}{16}\bigg[16(1) + 15(-1/3) + 10(1/5) + 3(-1/7)\bigg]\)
\(\displaystyle \pi \approx 3.1428...\)
You can apply this to 1 - 1/2 + 1/3 - 1/4 to estimate ln(2), or log(2), if you prefer writing it that way, to give it
to two correct rounded decimal places.
Examples of others that it may be used on:
\(\displaystyle 1/1^2 - 1/2^2 + 1/3^2 - 1/4^2\)
\(\displaystyle \dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{4}} \)
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