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Mathematics 24 Online
OpenStudy (anonymous):

How can I use a p-series comparison to show whether the sum of arctan(2n)/ n^2 from n=1 to n= infinite is convergent or divergent?

OpenStudy (anonymous):

The numerator (arctan(2n)) is bounded by pi/2, so we can say\[\sum_{k=1}^{\infty} \frac{\arctan(2k)}{k^2}\le \sum_{k=1}^{\infty} \frac{\pi/2}{k^2}<\infty\]

OpenStudy (anonymous):

I suppose you could put another step in if you want, pulling the constant (pi/2) out in front of the sum for illustrative purposes, but there isn't any mathematical requirement to do so.

OpenStudy (anonymous):

Ok, thank you very much!

OpenStudy (anonymous):

No sweat. Happy to help.

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