Calculate the geometric series? Please

enter image source here

1 Answer
Apr 25, 2018

This is not geometric, but using the Maclaurin series expansion for arctangent, we see its value is arctan(1)=π4.

Explanation:

A geometric series is generally in the form

n=0a(r)n where a is the first term and r is the common ratio shared by every single term.

We have the series

n=0(1)n2n+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)=n=0x2n+1(1)n2n+1

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

arctan(1)=n=012n+1(1)n2n+1

arctan(1)=n=0(1)n2n+1

Well, 12n+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)=π4