-4

Does it converge or diverge or we can't tell?

$$∑_{n=1}^{\infty}(-1)^n$$

Or is there simply no concrete answer?

Thanks in advance.

k1.M
  • 5,428
Chapi
  • 107

3 Answers3

3

Hint: If $\sum a_n$ converges, you must have $a_n \to 0$. Does this happen here?

Ivo Terek
  • 77,665
  • Yes, I do know the limit for an doesn't exist so it's not 0. But that doesn't make it diverge, right? If so, my question is how can we classify this series? – Chapi May 28 '15 at 22:58
  • $$ \left( \sum a_n \text{ converges } \implies a_n \to 0\right) \iff \left( a_n \not\to 0 \implies \sum a_n \text{ does not converge.}\right)$$ – Ivo Terek May 28 '15 at 23:02
  • 1
    @Chapi Its sequence of partial sums is often called "oscillating", which implies non-convergence. It varies from person to person whether oscillating behavior is considered divergence. – pjs36 May 28 '15 at 23:03
  • 1
    @pjs36 The definition of a divergent sequence is a sequence that does not converge. – Dan Robertson May 28 '15 at 23:14
  • @DanRobertson Thank you, that was the concept I was missing – Chapi May 28 '15 at 23:16
  • @DanRobertson Thanks, I must have mis-recalled a definition there. I thought I remembered some authors allowing "oscillation" as a third category, but I must be mistaken. – pjs36 May 28 '15 at 23:26
2

That depends on how you are planning to use that sum. I believe that there are physicist who will insist on the expression being a half, but mathematically there is no convergence. The partial sums are zero and one alternatingly, therefore no limit exists.

If you go for the physics approach you will get the value thus:

Let $S=\sum_{n=1}^\infty (-1)^n$

Then $(-1)S=S-1$ since the multiplication just shifts the sum by one. Therefore $S=\frac{1}{2}$. This is similiar to the way you determine the value of a geometric series, if you overlook the fact that $S$ is of course undefined in the first place. So if $S$ is to have any value it will be $\frac{1}{2}$ since that's the only value that exhibits the same properties as the Sum.

So if this sum occurs somewhere but you know it must be representative of some finite number because it represents something physical, it's $\frac{1}{2}$.

Matthias Kaul
  • 79
  • I like your initial answer, but some elaboration on why physicists will differ from others (possibly mathematicians) on the existence of the series value and what the value is equal to would be nice :). – Jubbles May 28 '15 at 23:18
  • 1
    maybe have a look at this for an in depth exploration of this type of sum – Matthias Kaul May 28 '15 at 23:38
  • @MatthiasKaul Thank you so much, such good answers on that question. I don't understand why everyone answered me like I had no idea what I was asking... Again very interesting answers in that question – Chapi May 29 '15 at 00:55
2

When we say a series converges, we mean that the sequence of partial sums approaches a limit. Let's consider the sequence of partial sums of your series: $$\begin{matrix}-1 & = & -1 \\ -1 + 1 &= &0 \\ -1 + 1 - 1 &= &-1\end{matrix}$$ and so on, so it keeps alternating from $-1$ to $0$. It therefore doesn't approach a unique value. The series therefore diverges.

Also, when you ask "is there no concrete answer", it's clear you're misunderstanding. A given series either converges or diverges, and there's no in-between. When we say a series diverges, we only mean that it doesn't converge.

wlad
  • 8,185