Calculate the geometric series? Please

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1 Answer
Apr 25, 2018

This is not geometric, but using the Maclaurin series expansion for arctangent, we see its value is arctan(1)=pi/4arctan(1)=π4.

Explanation:

A geometric series is generally in the form

sum_(n=0)^ooa(r)^nn=0a(r)n where aa is the first term and rr is the common ratio shared by every single term.

We have the series

sum_(n=0)^oo(-1)^n/(2n+1)n=0(1)n2n+1

As a result of the 2n+12n+1 in the denominator, this is not geometric.

Geometric series do not have the index nn 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)arctan(x)=n=0x2n+1(1)n2n+1

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

arctan(1)=sum_(n=0)^oo1^(2n+1)(-1)^n/(2n+1)arctan(1)=n=012n+1(1)n2n+1

arctan(1)=sum_(n=0)^oo(-1)^n/(2n+1)arctan(1)=n=0(1)n2n+1

Well, 1^(2n+1)=112n+1=1. 11 raised to any power returns itself.

So, that's why in the given summation, 11 is absent entirely.

Then, since our series is equal to arctan(1)arctan(1), its value is arctan(1)=pi/4arctan(1)=π4