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How do I check convergence for this? $$\sum _{n=1}^{\infty }\:\arctan\left(\frac{1}{n^2+n+1}\right)$$

I don't think I should be using the series integral test, since we haven't studied it at the lecture. I tried using other tests but I can't seem to figure out this one. Anyone have any hints/ideas?

Nex
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MikhaelM
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  • This is not a duplicate. Without computing the sum explicitly, you can simply use $\arctan \frac{1}{n^2+n+1}\sim \frac{1}{n^2}$, and since the series of $1/n^2$ is convergent, so is yours. – Jean-Claude Arbaut Nov 10 '15 at 17:48

1 Answers1

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HINT: $$\sum _{n=1}^{\infty }\:arctan\left(\frac{1}{n^2+n+1}\right)$$ $$=\sum _{n=1}^{\infty }\:arctan\left(\frac{(n+1)-n}{1+n(n+1)}\right)$$ $$=\sum _{n=1}^{\infty }\:\left[arctan\left(n+1\right)-arctan\left(n\right)\right]$$

SchrodingersCat
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