# Question about the convergence of $\sum_{n=1}^\infty (-1)^n$

This series obviously converges because $\sum_{n=1}^\infty (-1)^n = -1+1-1+1-1+1-1...=0$, but if the series $\sum_{n=1}^\infty a_n$ converges $\implies\lim{a_n}=0$, but in this case, $a_n=(-1)^n$ doesn't converge to 0 because $\lim{a_{2n}}=1$ and $\lim{a_{2n-1}}=-1\implies\lim{a_n}\nexists$

What am I missing here?

• There's a problem in the equal sign of $-1+1-1+1 \cdots = 0$.
– user99914
Sep 9 '17 at 10:47
• Obviously diverges, surely? Sep 9 '17 at 11:05
• It equals $\frac{1}{2}$ apparently. Check out Numberphile on YouTube and see the video on it. I haven't been taught this type of maths (? What does this mean $\longrightarrow \sum$ ? ) but I did some research and found that out. I hope it is useful. Sep 9 '17 at 11:13
• @user477343 No... please don't give out sources if you are not familiar with the math yourself. I like Numberphile, but some of their videos confuse those who are not familiar enough with the mathematics they are talking about. It does not equal $1/2$ in the traditional sense. In the traditional sense the series diverges.
– Eff
Sep 9 '17 at 11:16

This series obviously converges

No it doesn't. The definition of convergence is that it approaches some value. In this case it keeps alternating between 0 and -1; it is not approaching anything.

• "The definition of convergence is that it approaches some value." Could you clarify it? Sep 9 '17 at 12:14

You are wrong from the start. Why do you assert that$$-1+1-1+1-1+\cdots=0?$$ If you are adding a finite number of terms, then you will get $0$ half of the times and $-1$ the other half.

• But how can that be possible? I can admit it's true, but really, if I cancel every -1 with the next +1 till $\infty$, why it doesn't sum 0? Sep 9 '17 at 10:59
• @puradrogasincortar My answer covers this naive way of summing the series and why it is not the only way (or the best way) you can assign a value to the sum.
– A.M.
Sep 9 '17 at 11:04
• @puradrogasincortar Really? What if you add a $1$ to the left? Then the answer will be $1+0=1$, right?! But using your “if I cancel every $1$ with the next $1$”, the answer should be $0$ again. Sep 9 '17 at 11:26

This is Grandi's series, but with $-1$ added to the beginning. Its value depends on the axioms you are using.

If you consider the partial sums, they will alternate between $-1$ and $0$, and thus the series is divergent and has no value.

Alternatively, we can use Cesàro summation, which requires finding $$\lim_{k\to\infty}(\frac{\sum_{n=1}^k (-1)^n}{k})$$ This is the average of the partial sums, which will evaluate to $-1/2$ in this case. We can also reach this value by considering that the series is a geometric progression with $a = -1$, $r = -1$ and therefore infinite sum of $\frac{a}{1-r} =\frac{-1}{1+1} = -\frac{1}{2}$, though geometric series are usually only considered valid for $|r| < 1$, which is not the case here.

Additionally, note that naive manipulation of the series can lead to values of $-1$ or $0$, in what is called the Eilenberg–Mazur swindle: $$-1 + 1-1+1-1+1-\,... =(-1+1)+(-1+1)+(-1+1)+\,...=0+0+0+\,...=0$$ $$-1 + 1-1+1-1+\,... =-1+(1-1)+(1-1)+\,...=-1+0+0+0+\,...=-1$$

Thus, the series can have no sum, or sums of $-\frac{1}{2}$, $0$ or $-1$ depending on what axioms you are using and what purpose you are finding an infinite sum for.

• It should be said that the limit of the partial sums is the standard definition of the value of an infinite series. And it is what the theorem $$a_n \not\to 0 \quad\implies\quad \sum\limits a_n\text{ diverges}$$ is based on. So if you don't use the standard definition of convergence, you will also have to throw out a lot of theorems on convergence.
– Eff
Sep 9 '17 at 11:14
• You can apply the geometric series if you consider the Abel sum of the series, which moves it into the radius of convergence, then takes the limit to the boundary. Sep 9 '17 at 12:15
• $$\sum_{n=1}^\infty a_n\stackrel{\text{Abel}}=\lim_{x\to1^-}\sum_{n=1}^\infty a_nx^n$$In particular,$$\sum_{n=1}^\infty(-1)^n\stackrel{\text{Abel}}=\lim_{x\to1^-}\sum_{n=1}^\infty(-1)^nx^n=\lim_{x\to1^-}\frac{-x}{1+x}=-\frac12$$ Sep 9 '17 at 12:37