2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.