sum of the series?

Question

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

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

$$S=i-1-i+1+i-1-i+1.....$$ 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 ???

This "series" has been considered by Euler. Have a look at the very interesting document (eulerarchive.maa.org/hedi/HEDI-2006-06.pdf) — Jean Marie, Dec 27 '16 at 16:29
The Cesaro sums tend to $\frac12$ but the series itself is divergent. — Henricus V., Dec 27 '16 at 16:43
(ctd from @Henry) Otherwise said, it is convergent in the sense of Cesaro (https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation). This convergence is known to be weaker (thus more "tolerant") that usual convergence. — Jean Marie, Dec 27 '16 at 17:04

Simply Beautiful Art · Answer 1 · 2016-12-28T12:56:23.817

4

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.

edited Dec 28 '16 at 12:56

answered Dec 27 '16 at 16:31

Simply Beautiful Art

74,685

What is $\eta(0)$ ? – Jean Marie Dec 27 '16 at 16:32
@JeanMarie it is the dirichlet eta function. – Simply Beautiful Art Dec 27 '16 at 16:32
Thanks. I was knowing the $\zeta$ function but not the $\eta$ function – Jean Marie Dec 27 '16 at 16:37
1

@JeanMarie they are the same except one has alternating signs. – Simply Beautiful Art Dec 27 '16 at 17:18
1

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. – J. M. ain't a mathematician Dec 28 '16 at 10:39
@J.M.isn'tamathematician Sorry, was on phone earlier, so rather hard to do. Thanks for reminding me! – Simply Beautiful Art Dec 28 '16 at 12:50
yes what i did is not valid – Nebo Alex Dec 28 '16 at 13:32

score 4 · Answer 2 · answered Dec 27 '16 at 16:57

4

Look this(Terence Tao-Analysis I):]1

answered Dec 27 '16 at 16:57

Zuo

786

I think its better to provide a link and summarize the passage here. – Simply Beautiful Art Dec 28 '16 at 12:51
But isn't it beautiful @SimpleArt?? – Vidyanshu Mishra Dec 28 '16 at 12:54
@THELONEWOLF. Depends if you are on school wifi and all the images are blocked. Then you can't see anything here. – Simply Beautiful Art Dec 28 '16 at 12:57
ooh :) :) :) :) – Vidyanshu Mishra Dec 28 '16 at 12:58
@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. – Zuo Dec 28 '16 at 16:03
No problem. Cheers for the good find too! – Simply Beautiful Art Dec 28 '16 at 16:04

score 3 · Answer 3 · answered Dec 27 '16 at 16:25

3

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.

answered Dec 27 '16 at 16:25

Juniven Acapulco

9,654

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 – Nebo Alex Dec 27 '16 at 16:31
2

@Boris it has to exist first for that to be the case. – Simply Beautiful Art Dec 27 '16 at 16:31

score 2 · Answer 4 · answered Dec 27 '16 at 16:41

2

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 $0.5$.

answered Dec 27 '16 at 16:41

76david76

1,535

Please see Cesàro summation – Simply Beautiful Art Dec 28 '16 at 12:51

score 2 · Answer 5 · answered Dec 27 '16 at 16:42

2

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

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

answered Dec 27 '16 at 16:42

hamam_Abdallah

62,951

sum of the series?

5 Answers5