Can someone please explain leibniz rule to me,

[Dell]

according to leibniz, if i have a series (An) with an alternating sign, in order for the series to converge, i need either |An| to converge, or |An| to diverge and An=0(for n->infinity) and A(n)>A(n+1)

BUT in the following series where An= ((-1)^(n-1))*1/(n+100sin(n)) the series |An| diverges, lim An=0 BUT how can i prove that An>An+1, in fact i don't think that it is true since sin(n) goes from -1 to 1, so an could be An<An+1, YET this series converges(according to the answer in my book). can anyone see how this is possible??

i thought that maybe the rule for An>An+1 is also for (n->infinity) but then i get An=An+1 so the rule still doesn't hold.

can anyone please give me the exact rule here, as well as explaining why this is not working.for me

 

[lippyka]

? The rule that Leibniz stated is for when the terms of the series have alternating signs. In your example, the terms of the series do not have alternating signs. Therefore, the rule does not apply here. In order for your series to converge, you need to show that the limit of the sequence of absolute values of the terms is 0. To do this, you can use the Squeeze Theorem. This states that if you have three sequences of real numbers a(n), b(n), and c(n) such that a(n) ≤ b(n) ≤ c(n) for all n and lim a(n) = lim c(n) = L, then lim b(n) = L. In your example, you can let a(n) = 0, b(n) = |((-1)^(n-1))*1/(n+100sin(n))|, and c(n) = 1/(n+100). Then, since lim a(n) = lim c(n) = 0, you can use the Squeeze Theorem to prove that lim b(n) = 0, which shows that the sequence of absolute values converges to 0. Therefore, your series converges.