Divergence test for real series

Question

Problem: $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}n}{n+1}$$

So after using the series divergence test, since the limit does not equal 0, the series will diverge. At least I think I did the test right. How would this look as a proof though? Do I just show my work for the divergence test and then say "by the series divergence test this series diverges"? I just took knowledge from my calculus 2 class and applied it here. However, I don't remember going over a series divergence test in my analysis class, so I'm a little skeptical.

Answer 1

You can just apply the test directly:

$\displaystyle \sum \frac{(-1)^{n+1}n}{n+1}$ does not converge because $\displaystyle \frac{(-1)^{n+1}n}{n+1}$ does not limit to zero. $\qquad \blacksquare$

Your concern is valid for an analysis course; certainly you want every tool rigorously proven before putting it to use. However, it isn't too hard to prove the divergence test from "first principles". Recall that, if $\displaystyle \sum a_n$ converges, this means that its sequence of partial sums $\displaystyle s_k = \sum_{n=1}^k a_n$ converges. In other words, if the series converges to some $L \in \mathbb{R}$, then given any $\varepsilon > 0$, we have $|s_m - L| < \varepsilon$ for all sufficiently large $m$.

Now suppose that $a_n$ does not converge to $0$. This means we can find a $\delta > 0$ such that $|a_n| > \delta$ for arbitrarily large $n$. What implications does this have for the sequence $s_k$ of partial sums with regards to convergence?

Answer 2

Let $\sum_{n = 1}^\infty a_n$ a convergent series. Then, the sequence of summans $a_n$ is a null sequence, i.e. $\lim_{n \to \infty} a_n = 0$.

This gives you a necessary condition for a series to converge. The contraposition of this statement reads

If $a_n$ is not a null sequence, then $\sum_{n = 1}^\infty a_n$ cannot converge.

Now let's take a look at the sequence $a_n = \frac{(-1)^n n}{n + 1}$. This sequence does not converge at all (why?), i.e. in particular $a_n$ is not a null sequence. Thus by the second highlighted statement, the series $\sum_{n = 1}^\infty a_n$ cannot converge.