Find the value of \(\displaystyle{\sum_{{{n}={1}}}^{{{50}}}}{\arctan{{\left({\frac{{{2}{n}}}{{{n}^{{4}}-{n}^{{2}}+{1}}}}\right)}}}\)

George Michael

George Michael

Answered question

2022-04-08

Find the value of
n=150arctan(2nn4n2+1)

Answer & Explanation

Jameson Jensen

Jameson Jensen

Beginner2022-04-09Added 16 answers

Notice that we can write
2nn4n2+1=(n2+n)(n2n)1+(n2+n)(n2n)
In view of the addition formula for the tangent, we find that
arctan(xy1+xy)=arctanxarctany
for any x>y>0. Thus we have
arctan(2nn4n2+1)=arctan(n2+n)arctan(n2n)
Then you can easily compute the sum by telescoping.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?