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

  1. Geometric series: $\displaystyle \sum_{n\ge 0}2^{-n}$

    $\displaystyle 2^0 + 2^{-1} + 2^{-2} + 2^{-3} + \ldots = 2.$

  2. Telescoping series: $\displaystyle \sum_{n\ge 1}\frac{1}{n(n+1)}$

    $\displaystyle \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \frac{1}{4\cdot 5} + \ldots = 1.$

  3. James Gregory's (or Leibniz) series

    $\displaystyle \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{\pi}{4}$.

  4. Euler's series: $\displaystyle \sum_{n\ge 1} n^{-2}$

    $\displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi ^2}{6}.$

  5. Euler's series: $\displaystyle \sum_{n\ge 1} (2n-1)^{-2}$

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

  6. Euler's alternating series: $\displaystyle \sum_{n\ge 1}(-1)^{n+1}n^{-2}$

    $\displaystyle 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots = \frac{\pi ^2}{12}.$

  7. Euler's alternating series: $\displaystyle \sum_{n\ge 1}(-1)^{n+1}(2n-1)^{-3}$

    $\displaystyle 1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \frac{1}{9^3} - \frac{1}{11^3} + \cdots = \frac{\pi ^2}{32}.$

  8. Alternating Harmonic Series:

    $\displaystyle \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = ln(2)$.

  9. Nilakantha (15th century) I:

    $\displaystyle \frac{1}{1^5+4\cdot 1} - \frac{1}{3^5+4\cdot 3} + \frac{1}{5^5+4\cdot 5} - \frac{1}{7^5+4\cdot 7} + \ldots = \frac{\pi}{16}.$

  10. Nilakantha (15th century) II:

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

Infinite products

  1. John Wallis' product

    $\displaystyle \frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot \ldots}{1\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\cdot \ldots} = \frac{\pi}{2}.$

  2. François Viète's product

    $\displaystyle \frac{1}{2}\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2\sqrt{2}}}}}{2} \cdot \ldots = \frac{1}{\pi}.$

  3. $\displaystyle \prod_{n\ge 2}(1 - n^{-2})$

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

  4. $\displaystyle \prod_{n\ge 3}(1 - 4n^{-2})$

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

Continued fractions

  1. $\displaystyle \pi$ I

    $\displaystyle 1 + \frac{1^2}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+ \ldots}}}} = \frac{4}{\pi}.$

  2. $\displaystyle \pi$ II

    $\displaystyle 1 + \frac{1^2}{3+\frac{2^2}{5+\frac{3^2}{7+\frac{4^2}{9+ \ldots}}}} = \frac{4}{\pi}.$

  3. Golden ratio

    $\displaystyle 1 + \frac{1}{1+\frac{1}{1+\frac{1}{1+ \ldots}}} = \phi = \frac{1+\sqrt{5}}{2}.$

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

What not

  1. Golden ratio

    $\displaystyle \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}+\ldots}}} = \phi = \frac{1+\sqrt{5}}{2}.$

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

References

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

Related material
Read more...

Telescoping situations

  • 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
  • An Elementary Proof for Euler's Series
  • $\sin 1^{\circ}+\sin {2^\circ}+\sin 3^{\circ}+\cdots+\sin 180^{\circ}=\tan 89.5^{\circ}$
  • Problem 3824 from Crux Mathematicorum
  • $x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)$
  • A Welcome Problem for the Year 2018
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