We are given to evaluate:
Sn=n∑k=0[tan−1(1k2+k+1)]
My solution:
Using the identity:
tan−1(x)=cot−1(1x)
we may write:
Sn=n∑k=0[cot−1(k2+k+1)]
Now, using the fact that:
k2+k+1=k(k+1)+1(k+1)−k
and the identity:
cot(α−β)=cot(α)cot(β)+1cot(β)−cot(α)
We may now write:
Sn=n∑k=0[cot−1(k)−cot−1(k+1)]
This is a telescoping series, hence:
Sn=cot−1(0)−cot−1(n+1)=π2−cot−1(n+1)
Using the identity:
tan−1(x)+cot−1(x)=π2
We may also write:
Sn=tan−1(n+1)
And so we have found:
n∑k=0[tan−1(1k2+k+1)]=tan−1(n+1)
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