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# Determine the convergence of the series 1 – n sin (1/n)

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

1. Anonymous says:

This was already proved in exercise #17 when we proving absolute convergence of the same series but with an alternating sign.

2. Eiji says:

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.

• Eiji says:

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

• Anonymous says:

Yes thats a lot easier.