Expert Answer to "That is equivalent toπ=12∑n≥0(−1)n3n(2n+1)"

Olivia Vasquez

Answered question

2022-02-25

That is equivalent to
$π = \sqrt{12} \sum_{n \geq 0} \frac{(- 1)^{n}}{3^{n} (2 n + 1)}$

mtakadamu9i5

Beginner2022-02-26Added 8 answers

Another answer:

\frac{π}{\sqrt{12}} = \sqrt{3} \arctan (\frac{1}{\sqrt{3}})

= \sqrt{3} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1 + t^{2}} d t

= \sqrt{3} \int_{0}^{\frac{1}{\sqrt{3}}} \sum_{n = 0}^{\infty} {(- t^{2})}^{n} d t

= \sqrt{3} \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\frac{1}{\sqrt{3}}} t^{2 n} d t

= \sqrt{3} \sum_{n = 0}^{\infty} {(- 1)}^{n} {[\frac{t^{2 n + 1}}{2 n + 1}]}_{0}^{\frac{1}{\sqrt{3}}}

= \sqrt{3} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} ({(\frac{1}{\sqrt{3}})}^{2 n + 1} - 0)

= \frac{\sqrt{3}}{\sqrt{3}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(\sqrt{3})}^{2 n} (2 n + 1)}

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3^{n} (2 n + 1)}

π = \sqrt{12} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3^{n} (2 n + 1)}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?