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let sum of the series

$$S=1-1+1-1+1-1+1-1+1-1........$$

$S=\frac{1}{2}$

my question is if there are even number of terms then the sum is $0$

and if the number of terms are odd then sum is $1$.

but we don't know whether its odd or even because the sequence goes to infinity.

Is the answer derived by using the probability of getting either $0$ or $1$


I am making an edit to this, can it done by using the where $i $ is 4th root of unity

  1. $$S=i+i^2+i^3+i^4+i^5.....$$

$$S=i-1-i+1+i-1-i+1.....$$ then $1$ can be summed up using infinite sum of Gp which is $$S=\frac{a}{1-r}=\frac{i}{1-i}$$ by simplifying we get $$S=\frac{i}{1-i}\frac{1+i}{1+i}=\frac{i-1}{2}$$ and take the real part of $S=Re \big[ \frac{i-1}{2}\big]=-\frac{1}{2}$

is it a wrong way to solve this problem ???

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  • $\begingroup$ this series (infinite) has no sum $\endgroup$ Dec 27 '16 at 16:22
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    $\begingroup$ How do you define your sequence of points 1-1+.......... $\endgroup$ Dec 27 '16 at 16:24
  • $\begingroup$ Look up Grandi's series. $\endgroup$ Dec 27 '16 at 16:25
  • $\begingroup$ This "series" has been considered by Euler. Have a look at the very interesting document (eulerarchive.maa.org/hedi/HEDI-2006-06.pdf) $\endgroup$
    – Jean Marie
    Dec 27 '16 at 16:29
  • $\begingroup$ Sorry I should not use the proved statement @G.Sassatelli $\endgroup$
    – Nebo Alex
    Dec 27 '16 at 16:32
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Look this(Terence Tao-Analysis I):enter image description here[1]]1

enter image description here

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  • $\begingroup$ I think its better to provide a link and summarize the passage here. $\endgroup$ Dec 28 '16 at 12:51
  • $\begingroup$ But isn't it beautiful @SimpleArt?? $\endgroup$ Dec 28 '16 at 12:54
  • $\begingroup$ @THELONEWOLF. Depends if you are on school wifi and all the images are blocked. Then you can't see anything here. $\endgroup$ Dec 28 '16 at 12:57
  • $\begingroup$ ooh :) :) :) :) $\endgroup$ Dec 28 '16 at 12:58
  • $\begingroup$ @Simple Art,I get it.Because I want to show more content directly, but in my mobile phone will be very inconvenient to edit them. Thank you for your suggestion, next time I will pay attention. $\endgroup$
    – Zuo
    Dec 28 '16 at 16:03
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The solution is not "$1/2$", but rather it is regularized to give one half:

$$S=\lim_{x\ \uparrow\ 1}1-x+x^2-x^3+\dots=\frac12\\S=\eta(0)=\frac12\\S=\lim_{n\to\infty}\frac{S_n}n=\frac12$$

where $\eta(z)$ is the Dirichlet eta function and $S_n$ is the $n$th partial sum.

Other methods also yield $1/2$, but you are dealing with a divergent series, so any ideas like grouping do not make sense.


Specifically concerning your solution, you should notice that the real part of your series is

$$-1+1-1+1-1+\dots$$

which is $-1$ times the normal result. A more correct way to use your idea would be to take

$$S=\Re(\lim_{x\to i}1+x+x^2+x^3+x^4+\dots)$$

since the geometric series only works for $|x|<1$, we have to avoid the problem of convergence via limits. Notice that you cannot apply the limit to each term individually, as you will get a divergent series. Instead, you should calculate the series first, then take the limit, then the real part.

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  • $\begingroup$ What is $\eta(0)$ ? $\endgroup$
    – Jean Marie
    Dec 27 '16 at 16:32
  • $\begingroup$ @JeanMarie it is the dirichlet eta function. $\endgroup$ Dec 27 '16 at 16:32
  • $\begingroup$ Thanks. I was knowing the $\zeta$ function but not the $\eta$ function $\endgroup$
    – Jean Marie
    Dec 27 '16 at 16:37
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    $\begingroup$ @JeanMarie they are the same except one has alternating signs. $\endgroup$ Dec 27 '16 at 17:18
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    $\begingroup$ Jean's question is a hint that you should probably include a link to Dirichlet $\eta$ in your answer. OTOH, $\frac1{1-(-1)}$ (the geometric series) is more elementary. $\endgroup$ Dec 28 '16 at 10:39
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That series is $$\sum_{n=0}^{+\infty}(-1)^{n}$$ whic has no sum because $$\lim_{n\to +\infty}(-1)^n\neq 0.$$

So, writing as $S=1-1+1-1+\dots$ is not allowed. Hope you are clarified.

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  • $\begingroup$ why does summing $S=1-1+1-1+1-1.......$ ,$S=1-S$,$S=\frac{1}{2}$ ,is it not summation or its just letting that its equal to S $\endgroup$
    – Nebo Alex
    Dec 27 '16 at 16:31
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    $\begingroup$ @Boris it has to exist first for that to be the case. $\endgroup$ Dec 27 '16 at 16:31
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This is called Grandi's series, you can see about it here.

I think he defined it to be $0.5$ in some meaning the average of all its partial sums, is equal to $0.5$.

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For $n\geq 0$, let $$S_n=\sum_{k=0}^n (-1)^k.$$

then

$$S_{2n}=1$$ $$S_{2n+1}=0$$

$\implies (S_n)$ is not convergent. thus

$$\sum_{n=0}^{+\infty}(-1)^n\notin \Bbb R$$.

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