Leibniz and convergence?

[Dell]

in the following question i am given the series:

Σ((-1)^(n-1))*1/(n+100sin(n))

and am asked if the series converges or diverges.

as far as i know Leibniz's law for series with alternating signs states that if the series of the absolute values diverges then we check the following 2 conditions for convergence:
#1) lim(n->infinity) An =0
#2) An > A(n+1)

i have managed to prove that the absolute series diverges so now i need to check the 2 conditions,

#1) lim(n->infinity) An =0 can easily be proven so i move onto the next condition

#2) An > A(n+1)
since An is dependant on sin(n), how can i say for sure which is larger? since sin(n) can vary from -1 to 1, does this not mean that the series diverges??

in my book it says that the series converges, is this a misprint on their behalf or am i missing something here?

 

[lippyka]

The series does converge. The second condition of Leibniz's Law for Alternating Series states that the terms in the series must be decreasing, not necessarily greater than the next term. In this case, since the terms are approaching 0, the series converges.