If I have the series:
$$\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+2}$$
I need to know if this series converges or diverges. The solution says to use the n-th term test for divergence. In general I know that means that if you take $\lim_{n\rightarrow\infty} a_{n}$ and get any number other than 0, it diverges. However, I'm confused here because of the $(-1)^n$
The solution says to do
$\lim_{n\rightarrow\infty}\frac{n}{n+2}$ and this diverges by n-th term test since it equals $1$. However, can I just do this anytime I have something that has alternating $(-1)^n$? Just apply the n-th term test to the part that doesn't have the $(-1)^n$ and if that diverges then the whole thing diverges?
$\endgroup$ 23 Answers
$\begingroup$The subsequence $\left((-1)^{2n}\frac{2n}{2n+1}\right)_{n\in\Bbb N}$ of the sequence$$\left((-1)^n\frac n{n+1}\right)_{n\in\Bbb N}\tag1$$ is the sequence $\left(\frac{2n}{2n+1}\right)_{n\in\Bbb N}$ whose limit is $1$. Since the sequence $(1)$ has a subsequence which converges to a number different from $0$, it does not converge to $0$, and therefore your series diverges.
$\endgroup$ $\begingroup$Remember that the precise statement of this test is "If $\lim_{n\to\infty} a_n$ does not converge to $0$, then $\sum_{n=1}^\infty a_n$ diverges". That's broader than "If $\lim_{n\to\infty} a_n$ converges to a nonzero value" because it includes the possibility that $\lim_{n\to\infty} a_n$ doesn't exist, as is the case here.
$\endgroup$ $\begingroup$You can argue like this: If $(a_n)_{n}$ is not a nullsequence also $(-1)^n a_n$ is not a null sequence and hence the sequence is diverging.
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