Evaluate: \sum_{n=1}^m\arctan(\frac{1}{n^2+n+1})

Jamelia Daniels

Jamelia Daniels

Answered question

2022-03-01

Evaluate:
n=1marctan(1n2+n+1)

Answer & Explanation

paralovut91

paralovut91

Beginner2022-03-02Added 7 answers

This telescopes, using the fact that arctan(u)arctan(v)=arctan(uv1+uv)
Specifically take u=n+1 and b=n. Then
arctan(1n2+n+1)=arctan(n+1)arctan(n)
This gives
n=1marctan(1n2+n+1)=arctan(m+1)π4

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