Consider the series

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

The series converges absolutely.

*Proof.* We know from the previous exercise (Section 10.20, Exercise #30) that the series

converges absolutely. Using the limit comparison test we have

Now we consider these as functions of a real-variable and make the substitution , and use L’Hopital’s rule three times to take the limit,

Therefore, since converges absolutely we have established the absolute convergence of

You could also solve it using Taylor’s Polynomial. 1 – n(sin (1/n)) = 1 – n(1/n – ((1/n)^3)/3! + …) > ((1/n)^2)/6. Since we know that ((1/n)^2)/6 converges, 1 – n(sin (1/n)) converges.

I meant 1 – n(sin (1/n)) = 1 – n(1/n – ((1/n)^3)/3! + …) < ((1/n)^2)/6.