How do you tell if series tan(n) and series e^(-2/n) converge/diverge using the divergence test?

Question

Hello! I can't figure out whether series An = tan(n) or series An = e^-2/n converge or diverge using the divergence test. I know that tanx doesn't approach anything, but I'm also confused on that. Why doesn't An = tan(n) converge to pi/2?

Thank you for the help!

Answer 1

We need to make a very important distinction here between sequence and series:

The sequence is just the list of terms themselves. So, for example, the harmonic sequence a_n = 1/n is the list of terms 1 , 1/2 , 1/3 , 1/4 , ... It is trivial to show that this sequence converges to a limit, and that limit is 0.

A series is the sum of some, or all, of the terms in a sequence. So, for example, the harmonic series is

1 + 1/2 + 1/3 + 1/4 + .... It is more difficult, but possible, to show that this series diverges. Using formal mathematical notation, we write lim_n→∞ ∑₁ⁿ 1/k does not exist. Contrarily, the geometric series with a common ratio of 1/2, generated by the sequence 1 , 1/2 , 1/4 , 1/8 , ... , does converge to a limit of 2. Again, using correct notation, we write lim_n→∞ ∑₁ⁿ (1/2)^k-1 = 2. In other words, 1 + 1/2 + 1/4 + 1/8 + ... = 2.

Of central importance is this fact re sequences and series: the terms in the sequence converging to 0 is a necessary but not sufficient condition for the series (the sum of those terms) to converge. In other words, if lim_n→∞ a_n ≠ 0 , then the related series diverges. If lim_n→∞ a_n = 0 , then the related series may or may not converge, and other tests are required to determine which is the case.

None of the series you reference in your question converge, since none of the sequences converge to 0. However, if your question is asking instead about the sequences, then one converges, one doesn't.

The sequence a_n = tan(n) diverges because the tangent function is periodic and achieves y-values that range from - ∞ to ∞.

You may be confusing that with a_n = tan^-1(n) which converges to π/2 as n→∞. (The graph of the arctangent function has a horizontal asymptote at y = π/2 on the right-hand side of its graph, i.e. as x→∞.)

The other sequence, a_n = e^-2/n converges to 1, since lim_n→∞ - 2/n = 0 and e⁰ = 1.