What can we say about the series: $-1+1-1+1-1+1-1+\cdots=\sum_{n=1}^{\infty}(-1)^n$?

Question

What can we say about this series? $$-1+1-1+1-1+1-1+\cdots=\sum_{n=1}^{\infty}(-1)^n$$

Intuitively, the sum of it seems to converge to zero, as each term cancels the one before it, although none of the convergence tests seem to work on the series. Also I have no idea as for how to calculate the limit $\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty}(-1)^n$, which would have enabled me to at least some degree to determine what is going to happen with this sum.

Answer 1

Since the limit $\lim_{n\to\infty}(-1)^n$ doesn't exist, then, in particular, it is not $0$, and therefore the series diverges.

Answer 2

For the series to convergence the sequence within would have to be convergent to zero. Since it's not the series does not converge.

Answer 3

It's divergent, and it oscillates between two numbers, you cannot find a limit of such a series.

Answer 4

As you and the other answerers know, this series does has no sum when you use the standard first definition of convergence as the limit (if it exists) of the partial sums.

There are fancier ways to think about summing series. You can start reading here:

https://www.encyclopediaofmath.org/index.php/Summation_methods\

and here

https://en.wikipedia.org/wiki/Divergent_series

Answer 5

Well, using the standard definition of convergence, we can say about this series that it does not converge. This is easily seen when looking directly at the partial sums: $$\begin{align} \sum_{n=1}^1 (-1)^n &= -1\\ \sum_{n=1}^2 (-1)^n &= 0\\ \sum_{n=1}^3 (-1)^n &= -1\\ \sum_{n=1}^4 (-1)^n &= 0\\ &\ldots \end{align}$$ Obviously that sequence never remains close to any value.

Now for any non-standard definition of convergence (and there are many of them), as long as they fulfill a few reasonable conditions, you can say:

If the series converges according to some criterion, its value has to be $-\frac12$:

We have $$\begin{align} \sum_{n=1}^{\infty} (-1)^n &= -1 + \sum_{n=2}^{\infty}(-1)^n\\ &= -1 + \sum_{n=1}^{\infty}(-1)^{n+1}\\ &= -1 - \sum_{n=1}^{\infty} (-1)^n \end{align}$$ and therefore $$2 \sum_{n=1}^{\infty} (-1)^n = -1.$$

Answer 6

For these kind of not convergent infinite series we can assign values by Cesàro sum, defined as the limit as $n\to \infty$ of the sequence of arithmetic means of the first $n$ partial sums of the series.

In this sense we can assign to the series (aka Grandi's series) the value $\frac 12$.

Answer 7

It is divergent so it does not have a value ,but one can obtain it to be $-\frac12$. Correct me if i'm wrong below

Proof:

Consider : $S =-1+1-1+1-1+1-1+1-1+...........$

Now ,add brackets :

$S = (-1+1)-(-1+1)-(-1+1)-.......=0$

One can change the order of the brackets without changing the value:

$S = -1 +(1-1)+(1-1) +(1-1).......= -1$

add the two, giving $2S = -1 \implies S =-\frac12$

A closely related series is the Grandi's series which is $\sum_{i=0}^\infty (-1)^i$ which evaluates out to $\frac12$

Answer 8

Just take a look at the definition of a limit. We say that:

$$\lim_{n\rightarrow\infty}a_n=a$$

IF for any $\varepsilon>0$, there is an $N$ such that for all $n>N$, we have: $$|a_n-a|<\varepsilon$$

This is not the case for $0$, $1$ or $-1$.