So I've been practicing several series that can be solved using the alternating series test, but I've came to a question that's been bothering me for sometime now.(adsbygoogle = window.adsbygoogle || []).push({});

If a series fails the alternating series test, will the test for divergence always prove it to be divergent?

Typically, in most examples that I find on the James Stewart website they completely omit the alternating component when using the test for divergence. Is this because it will make the limit not exist?

Some examples.

Ʃ n = 1 to ∞ {(-1)^n * 2^(1/n)}

The alternating series test fails, so in the solution they take the

lim n → ∞ 2^(1/n) = 1.

Ʃ n = 1 to ∞ {(-1)^n * ((2^n) / n^2))}

Similarly, the alternating series test fails, so they use the test for divergence.

Again, they fail to include the (-1)^n and conclude that

lim n → ∞ 2^n / n^2 = ∞ ∴ the series is divergent.

Why do they not include the (-1)^n, wont this make the limit not exist? Obviously, this will still prove the series to diverge, but which one is the correct way to do it? Should I write the limit does not exist or the limit = 1? Thanks in advance to anyone that can help me out!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Alternating Series Test/Test for Divergence

**Physics Forums | Science Articles, Homework Help, Discussion**