Lim 1/(arctan(2n)) for n ---> infinity = 2/pi.
So, because the limit is not zero, the series cannot converge. Note that a limit of zero does not guarantee convergence, so it is a necessary but not sufficient condition for convergence that, in this case, was not met.
one last one, so sorry! thanks so much!
how would I test the series for convergence or divergence?
the series from n=1 to infinity of 1/(arctan(2n))
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