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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?

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    $\begingroup$ There's a problem in the equal sign of $-1+1-1+1 \cdots = 0$. $\endgroup$
    – user99914
    Sep 9 '17 at 10:47
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    $\begingroup$ Obviously diverges, surely? $\endgroup$ Sep 9 '17 at 11:05
  • $\begingroup$ 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. $\endgroup$
    – Mr Pie
    Sep 9 '17 at 11:13
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    $\begingroup$ @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. $\endgroup$
    – Eff
    Sep 9 '17 at 11:16
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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.

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

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  • $\begingroup$ 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? $\endgroup$ Sep 9 '17 at 10:59
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    $\begingroup$ @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. $\endgroup$
    – A.M.
    Sep 9 '17 at 11:04
  • $\begingroup$ @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. $\endgroup$ Sep 9 '17 at 11:26
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Eff
    Sep 9 '17 at 11:14
  • $\begingroup$ 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. $\endgroup$ Sep 9 '17 at 12:15
  • $\begingroup$ $$\sum_{n=1}^\infty a_n\stackrel{\text{Abel}}=\lim_{x\to1^-}\sum_{n=1}^\infty a_nx^n$$In particular,$$\sum_{n=1}^\infty(-1)^n\stackrel{\text{Abel}}=\lim_{x\to1^-}\sum_{n=1}^\infty(-1)^nx^n=\lim_{x\to1^-}\frac{-x}{1+x}=-\frac12$$ $\endgroup$ Sep 9 '17 at 12:37

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