test the series for convergence or divergence
the series from n=1 to infinity of
1/(arctan(2n))
I again didn't know what test to use
To test the convergence or divergence of a series, we can use the Comparison Test. We compare the given series to a known series whose convergence or divergence is already determined.
In this case, let's use the Comparison Test with the harmonic series, which is known to diverge:
1/n ≥ 1/(arctan(2n)) [Note: We replaced arctan(2n) with 1 for comparison]
Now, let's take the limit as n approaches infinity for both sides:
lim(n→∞) (1/n) ≥ lim(n→∞) (1/(arctan(2n)))
The limit of 1/n as n approaches infinity is 0, and since arctan(2n) is always positive, we have:
0 ≥ lim(n→∞) (1/(arctan(2n)))
Since the right-hand side is positive, the left-hand side must be as well. Therefore, the comparison implies that the given series also diverges.
So, the series from n=1 to infinity of 1/(arctan(2n)) is a divergent series.