Series

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.2 – Series

Series

Series to Compute $\pi$$\pi$

NOTE: You will NOT be assessed on any of the following series.

The Madhava of Sangamagrama Series

The Madhava of Sangamagrama (c.1350--c.1425) [in the West, the series is known as the Liebniz-Gregory (c.1676)] Series:

${\displaystyle \frac{\pi }{4}=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{1}{2k+1}}$$\displaystyle\frac{\pi}{4}=\sum_{k=0}^{\infty}{(-1)^k\frac{1}{2k+1}}$

Ramanujan

In 1914, Ramanujan identified the following:

${\displaystyle \frac{1}{\pi }=\frac{2\sqrt{2}}{9801}\sum _{k=0}^{\mathrm{\infty }}\frac{(4k)!(1103+26390k)}{(k!{)}^4{396}^{4k}}}$$\displaystyle\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{k=0}^{\infty}{\frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}}$

and

${\displaystyle \frac{4}{\pi }=\frac{1}{882}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{(4k)!(1123+21460k)}{(4^{k}k!{)}^4{882}^{2k}}}$$\displaystyle\frac{4}{\pi}=\frac{1}{882}\sum_{k=0}^{\infty}{(-1)^k\frac{(4k)!(1123+21460k)}{(4^k k!)^4 882^{2k}}}$

Chudonovsky

In 1987, Chudonovsky identified the following:

${\displaystyle \frac{1}{\pi }=\frac{1}{53360\sqrt{640320}}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{(6k)!(13591409+545140134k)}{(k!{)}^3(3k)!{640320}^{3k}}}$$\displaystyle\frac{1}{\pi}=\frac{1}{53360\sqrt{640320}}\sum_{k=0}^{\infty}{(-1)^k\frac{(6k)!(13591409+545140134k)}{(k!)^3 (3k)! 640320^{3k}}}$

Bailey, Borwein and Plouffe

In 1985, Bailey, Borwein, and Plouffe identified the following:

${\displaystyle \pi =\sum _{k=0}^{\mathrm{\infty }}(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6})\frac{1}{{16}^k}}$$\displaystyle\pi = \sum_{k=0}^{\infty}{\left(\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right)\frac{1}{16^k}}$

Roger Ap'{e}ry identified the following:

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}}$$\displaystyle\sum_{k=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}$

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{n^4}=\frac{{\pi }^4}{90}}$$\displaystyle\sum_{k=1}^{\infty}{\frac{1}{n^4}}=\frac{\pi^4}{90}$

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{n^6}=\frac{{\pi }^6}{945}}$$\displaystyle\sum_{k=1}^{\infty}{\frac{1}{n^6}}=\frac{\pi^6}{945}$

and others, $\dots$$\ldots$

Machin (1873) identified the following:

$\begin{array}{c}{\displaystyle \frac{\pi }{4}=4{\tan }^{-1}\left(\frac{1}{5}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)}\\ =4[\frac{1}{5}-\frac{1}{3}{\left(\frac{1}{5}\right)}^3+\frac{1}{5}{\left(\frac{1}{5}\right)}^5-\frac{1}{7}{\left(\frac{1}{5}\right)}^7+\cdots ]-[\frac{1}{239}-\frac{1}{3}{\left(\frac{1}{239}\right)}^3+\frac{1}{5}{\left(\frac{1}{239}\right)}^5-\frac{1}{7}{\left(\frac{1}{239}\right)}^7+\cdots ]\end{array}$$\displaystyle\frac{\pi}{4} = 4\tan^{-1}{\left(\frac{1}{5}\right)}-\tan^{-1}{\left(\frac{1}{239}\right)} \\ = 4\left[\frac{1}{5}-\frac{1}{3}\left(\frac{1}{5}\right)^3+\frac{1}{5}\left(\frac{1}{5}\right)^5-\frac{1}{7}\left(\frac{1}{5}\right)^7+\cdots\right] - \left[\frac{1}{239}-\frac{1}{3}\left(\frac{1}{239}\right)^3+\frac{1}{5}\left(\frac{1}{239}\right)^5-\frac{1}{7}\left(\frac{1}{239}\right)^7+\cdots\right]$

Additional Series Involving $\pi$$\pi$

${\displaystyle \frac{\pi \sqrt{2}}{4}=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{13}-\frac{1}{15}+\cdots }$$\displaystyle\frac{\pi\sqrt{2}}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{13} - \frac{1}{15} + \cdots$

${\displaystyle \frac{\pi \sqrt{3}}{6}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\cdots }$$\displaystyle\frac{\pi\sqrt{3}}{6} = 1 - \frac{1}{5} + \frac{1}{7} - \frac{1}{11} + \frac{1}{13} - \frac{1}{17} + \cdots$

${\displaystyle \frac{\pi \sqrt{3}}{9}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{8}+\cdots }$$\displaystyle\frac{\pi\sqrt{3}}{9} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{5} + \frac{1}{7} - \frac{1}{8} + \cdots$

${\displaystyle \frac{{\pi }^2}{8}=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(2k+1{)}^2}}$$\displaystyle\frac{\pi^2}{8} = \sum_{k=0}^{\infty}{\frac{1}{(2k+1)^2}}$

${\displaystyle \frac{{\pi }^4}{96}=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(2k+1{)}^4}}$$\displaystyle\frac{\pi^4}{96} = \sum_{k=0}^{\infty}{\frac{1}{(2k+1)^4}}$

${\displaystyle \frac{{\pi }^6}{960}=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(2k+1{)}^6}}$$\displaystyle\frac{\pi^6}{960} = \sum_{k=0}^{\infty}{\frac{1}{(2k+1)^6}}$

${\displaystyle \frac{{\pi }^3}{32}=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{1}{(2k+1{)}^3}}$$\displaystyle\frac{\pi^3}{32} = \sum_{k=0}^{\infty}{(-1)^k\frac{1}{(2k+1)^3}}$

${\displaystyle \frac{5{\pi }^5}{1536}=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{1}{(2k+1{)}^5}}$$\displaystyle\frac{5\pi^5}{1536} = \sum_{k=0}^{\infty}{(-1)^k\frac{1}{(2k+1)^5}}$

${\displaystyle \frac{61{\pi }^7}{184320}=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{1}{(2k+1{)}^7}}$$\displaystyle\frac{61\pi^7}{184320} = \sum_{k=0}^{\infty}{(-1)^k\frac{1}{(2k+1)^7}}$

${\displaystyle \frac{{\pi }^2}{12}=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1}{k^2}}$$\displaystyle\frac{\pi^2}{12} = \sum_{k=1}^{\infty}{(-1)^{k+1}\frac{1}{k^2}}$

${\displaystyle \frac{7{\pi }^4}{720}=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1}{k^4}}$$\displaystyle\frac{7\pi^4}{720} = \sum_{k=1}^{\infty}{(-1)^{k+1}\frac{1}{k^4}}$

${\displaystyle \frac{31{\pi }^4}{30240}=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1}{k^6}}$$\displaystyle\frac{31\pi^4}{30240} = \sum_{k=1}^{\infty}{(-1)^{k+1}\frac{1}{k^6}}$

${\displaystyle \frac{{\pi }^2}{20}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{{11}^2}+\cdots }$$\displaystyle\frac{\pi^2}{20} = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{11^2} +\cdots$

${\displaystyle \frac{{\pi }^4}{1260}=\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{5^4}+\frac{1}{7^4}+\frac{1}{8^4}+\frac{1}{{11}^4}+\cdots }$$\displaystyle\frac{\pi^4}{1260} = \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{5^4} + \frac{1}{7^4} + \frac{1}{8^4} + \frac{1}{11^4} +\cdots$

${\displaystyle \frac{4{\pi }^6}{225225}=\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{5^6}+\frac{1}{7^6}+\frac{1}{8^6}+\frac{1}{{11}^6}+\cdots }$$\displaystyle\frac{4\pi^6}{225225} = \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{5^6} + \frac{1}{7^6} + \frac{1}{8^6} + \frac{1}{11^6} +\cdots$

Euler's Formula

NOTE: You will NOT be assessed on Euler’s Formula.

${\displaystyle e^{ix}=\cos (x)+i\sin (x)}$$\displaystyle e^{ix}=\cos{(x)} + i\sin{(x)}$

where $\sqrt{-1}=i$$\sqrt{-1}=i$

Euler’s Formula is important for understanding solutions to higher-order ODE that have characteristic equations with complex roots.

When $x=\pi$$x=\pi$, then $e^{i\pi }=-1$$e^{i\pi}=-1$ which can be proven using Maclaurin Series for $f(x)=e^x$$f(x)=e^x$, $f(x)=\sin (x)$$f(x)=\sin{(x)}$, and $f(x)=\cos (x)$$f(x)=\cos{(x)}$ [See CPT_25].

NOTE: $\cos (x)+i\sin (x)$$\cos{(x)} + i\sin{(x)}$ is often abbreviated as $\text{cis}(x)$$\text{cis}{(x)}$.

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Last Modified: Tuesday, 1 December 2020 15:47 EDT