# Calculate the geometric series? Please

Apr 25, 2018

This is not geometric, but using the Maclaurin series expansion for arctangent, we see its value is $\arctan \left(1\right) = \frac{\pi}{4}$.

#### Explanation:

A geometric series is generally in the form

${\sum}_{n = 0}^{\infty} a {\left(r\right)}^{n}$ where $a$ is the first term and $r$ is the common ratio shared by every single term.

We have the series

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right)$

As a result of the $2 n + 1$ in the denominator, this is not geometric.

Geometric series do not have the index $n$ anywhere but the exponent.

However, we can still find the exact value of this series.

Recall the following Maclaurin series expansion for the arctangent:

$\arctan \left(x\right) = {\sum}_{n = 0}^{\infty} {x}^{2 n + 1} {\left(- 1\right)}^{n} / \left(2 n + 1\right)$

For our series, we can see that $x = 1 ,$ so we really have

$\arctan \left(1\right) = {\sum}_{n = 0}^{\infty} {1}^{2 n + 1} {\left(- 1\right)}^{n} / \left(2 n + 1\right)$

$\arctan \left(1\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right)$

Well, ${1}^{2 n + 1} = 1$. $1$ raised to any power returns itself.

So, that's why in the given summation, $1$ is absent entirely.

Then, since our series is equal to $\arctan \left(1\right)$, its value is $\arctan \left(1\right) = \frac{\pi}{4}$