# A necessary condition for convergence of an alternate series

I was doing some exercises about alternate series and at some of them i could show that that the series was not absolutely convergent, since the term $a_n \not\rightarrow 0$ as $n \rightarrow \infty$ .

So i came up with the question: is $\lim a_n = 0$ also a necessary condition for the convergence of the alternating series $\sum (-1)^na_n?$

Here's my anwser:

Suppose that $\sum (-1)^na_n$ has a finite sum $s$. Let $s_n=a_1-a_2+...+(-1)^{n-1}a_n$ be the sequence of the partial sums. We have that $\lim s_n = s$ and also that $\lim s_{n-1} = s$, since $(s_{n-1})$ is a subsequence of $(s_n)$ . Therefore,

$$0 = s-s = \lim s_n - \lim s_{n-1} = \lim (s_n - s_{n-1}) = \lim(-1)^{n-1}a_n$$

and this shows that $\lim a_n = 0$, since both even and odd subsequences of $b_n = (-1)^{n-1}a_n$ goes to the same limit: $0$.

Is it right?

• The condition $\lim_{n\to\infty} a_n = 0$ is a necessary condition for the convergence of any infinite series. It's called the "$n$-th term test (for divergence)" (i.e., if it fails, the series diverges). See:$\;$en.wikipedia.org/wiki/Term_test – quasi May 17 '17 at 18:31

Looks good - the only thing I was wary of was this statement: $\lim a_n - \lim b_n = \lim(a_n-b_n)$
But it looks like it is true if $\lim a_n=\lim b_n:=a$ exists.
Because: $$\forall\varepsilon>0\quad \exists N\in\mathbb{N}: |a_n-a|<\varepsilon\quad \forall n\ge N$$ then:
$$|a_n-b_n|=|a_n-a+a-b_n|\le|a_n-a|+|a-b_n|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\quad \forall n\ge N$$
So $\lim(a_n-b_n)=0$