Test the series for convergence or divergence.
Sum from 1 to infinity of tan(1/n)
Test for Divergence doesn't work, since we have tan(1/n) = 0.
I've considered the integral test, but the related function f(x) = tan(1/x) is discontinuous on the interval [1, infinity).
Root test doesn't look like an option, and I don't think the Limit Comparison test will help, because the form will be: tan(1/(1+x))/tan(1/x) since I don't know what I would do with that.
So I ran out of tests, and there has to be a way. Can anybody please help me?
Thank you.
Sum from 1 to infinity of tan(1/n)
Test for Divergence doesn't work, since we have tan(1/n) = 0.
I've considered the integral test, but the related function f(x) = tan(1/x) is discontinuous on the interval [1, infinity).
Root test doesn't look like an option, and I don't think the Limit Comparison test will help, because the form will be: tan(1/(1+x))/tan(1/x) since I don't know what I would do with that.
So I ran out of tests, and there has to be a way. Can anybody please help me?
Thank you.
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Use the limit comparison test with the harmonic series.
lim(n→∞) tan(1/n) / (1/n)
= lim(n→∞) (-1/n^2) sec^2(1/n) / (-1/n^2), by L'Hopital's Rule
= lim(n→∞) sec^2(1/n)
= sec^2(0)
= 1.
Since the harmonic series diverges, so does the series in question.
I hope this helps!
lim(n→∞) tan(1/n) / (1/n)
= lim(n→∞) (-1/n^2) sec^2(1/n) / (-1/n^2), by L'Hopital's Rule
= lim(n→∞) sec^2(1/n)
= sec^2(0)
= 1.
Since the harmonic series diverges, so does the series in question.
I hope this helps!