Infinite Sums and Products

Infinity, as an informal concept, is associated with endless repetition. Common expressions like "and so on" and "etcetera" and, occasionally, the ellipses "..." reflect a concept that mathematics attempts to make rigorous. On this page I shall collect a few appealing formulas whose meaning I hope will be intuitively clear even without formal justification. I believe this could be a good way to illustrate one of the uses of infinity in mathematics.

Infinite sums

Geometric series: \(\sum_{n\ge 0}2^{-n}\)

\(2^0 + 2^{-1} + 2^{-2} + 2^{-3} + \ldots = 2.\)
Telescoping series: \(\sum_{n\ge 1}\frac{1}{n(n+1)}\)

\(\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \frac{1}{4\cdot 5} + \ldots = 1.\)
James Gregory's (or Leibniz) series

\(\frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{\pi}{4}\).
Euler's series: \(\sum_{n\ge 1} n^{-2}\)

\(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi ^2}{6}.\)
Euler's series: \(\sum_{n\ge 1} (2n-1)^{-2}\)

\(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots = \frac{\pi ^2}{8}.\)
Euler's alternating series: \(\sum_{n\ge 1}(-1)^{n+1}n^{-2}\)

\(1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots = \frac{\pi ^2}{12}.\)
Euler's alternating series: \(\sum_{n\ge 1}(-1)^{n+1}(2n-1)^{-3}\)

\(1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \frac{1}{9^3} - \frac{1}{11^3} + \cdots = \frac{\pi ^2}{32}.\)
Alternating Harmonic Series:

\(\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = ln(2)\).
Nilakantha (15th century) I:

\(\frac{1}{1^5+4\cdot 1} - \frac{1}{3^5+4\cdot 3} + \frac{1}{5^5+4\cdot 5} - \frac{1}{7^5+4\cdot 7} + \ldots = \frac{\pi}{16}.\)
Nilakantha (15th century) II:

\(3+\frac{4}{3^3-3} - \frac{4}{5^3-5} + \frac{4}{7^3-7} - \frac{4}{9^3-9} + \ldots = \pi.\)

Infinite products

John Wallis' product

\(\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot \ldots}{1\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\cdot \ldots} = \frac{\pi}{2}.\)
François Viète's product

\(\frac{1}{2}\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2\sqrt{2}}}}}{2} \cdot \ldots = \frac{1}{\pi}.\)
\(\prod_{n\ge 2}(1 - n^{-2})\)

\((1 - \frac{1}{2^2})\cdot (1 - \frac{1}{3^2})\cdot (1 - \frac{1}{4^2})\cdot \ldots = \frac{1}{2}.\)
\(\prod_{n\ge 3}(1 - 4n^{-2})\)

\((1 - \frac{4}{3^2})\cdot (1 - \frac{4}{4^2})\cdot (1 - \frac{4}{5^2})\cdot \ldots = \frac{1}{6}.\)

Continued fractions

\(\pi\) I

\(1 + \frac{1^2}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+ \ldots}}}} = \frac{4}{\pi}.\)
\(\pi\) II

\(1 + \frac{1^2}{3+\frac{2^2}{5+\frac{3^2}{7+\frac{4^2}{9+ \ldots}}}} = \frac{4}{\pi}.\)
Golden ratio

\(1 + \frac{1}{1+\frac{1}{1+\frac{1}{1+ \ldots}}} = \phi = \frac{1+\sqrt{5}}{2}.\)

(For more, check the page on continued fractions.)

What not

Golden ratio

\(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}+\ldots}}} = \phi = \frac{1+\sqrt{5}}{2}.\)

(For more, check the page on continued fractions.)

References

J. Arndt, C. Haenel, π Unleashed, Springer, 2000

Related material Read more...

	Leibniz and Pascal Triangles
	Sum of an infinite series
	Harmonic Series And Its Parts
	A Telescoping Series
	An Inequality With an Infinite Series
	That Divergent Harmonic Series

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