Question

Calculate the geometric series? Please

Answer 1

This is not geometric, but using the Maclaurin series expansion for arctangent, we see its value is `arctan(1)=pi/4`.

Explanation:

A geometric series is generally in the form

`sum_(n=0)^ooa(r)^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)^oo(-1)^n/(2n+1)`

As a result of the `2n+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(x)=sum_(n=0)^oox^(2n+1)(-1)^n/(2n+1)`

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

`arctan(1)=sum_(n=0)^oo1^(2n+1)(-1)^n/(2n+1)`

`arctan(1)=sum_(n=0)^oo(-1)^n/(2n+1)`

Well, `1^(2n+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(1)`, its value is `arctan(1)=pi/4`