Calculate \sum_{n=1}^{\infty} \arctan (\frac{2\sqrt2}{n^2+1})

Babenzgs

Babenzgs

Answered question

2022-01-29

Calculate n=1arctan(22n2+1)

Answer & Explanation

ebbonxah

ebbonxah

Beginner2022-01-30Added 15 answers

Note that
arctan(22k2+1)=arctan((k+12)(k12)(k+12)(k12t)+1)=arctan(k+12)arctan(k12)
Therefore,
k=1narctan(22k2+1)=arctan(n+12)+arctan(n2)arctan(12)arctan(02)
Ergo,
k=1arctan(22k2+1)=πarctan(12)2.52611

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