Calculate the geometric series? Please

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Answer 1 · 2018-04-25T01:52:25

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

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#