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# Determine the convergence of the series sin (nπ + 1/log n)

Consider the series Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series is conditionally convergent.

Proof. First, we recall the formula for the sine of a sum, Therefore, we have (Here we use that for all and for all .) This series is convergent since and it is monotonically decreasing for .

This convergence is conditional since But then, since and is decreasing for we have for all . We know diverges since it is asymptotically equivalent to (i.e., ). Therefore, by the comparison test diverges. Therefore, the convergence of the series in question is conditional
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