Sum of series: \(\displaystyle{\arctan{{\left({n}+{1}\right)}}}-{\arctan{{\left({n}+{2}\right)}}}\) where n varies from

nastupnat0hh

nastupnat0hh

Answered question

2022-04-02

Sum of series:
arctan(n+1)arctan(n+2) where n varies from 0 to infinity.

Answer & Explanation

Regan Gallegos

Regan Gallegos

Beginner2022-04-03Added 9 answers

If you are looking for the sum of the series
n=0(arctan(n+1)arctan(n+2))
then what you have is a telescoping series. This means that terms will cancel. The first part of the sum is
arctan(1)arctan(2)+arctan(2)arctan(3)+arctan(3)arctan(4)+
The terms in the middle will cancel, for instance:
arctan(2)+arctan(2)=0
Thus your n-th partial sum is
Sn=arctan(1)arctan(n+2)
taking the limit gives
limn(arctan(1)arctan(n+2))=arctan(1)π2
And arctan(1)=π4

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