Does Grandi's series diverge or is it equal to $\frac{1}{2}$? Grandi's series :
$$\sum_{n=0}^\infty (-1)^n$$
I heard many many times about Grandi's series and also read about it in this site . In some sources , It's equal to $\frac{1}{2}$ and in the other it's undefined! I'm confused ! I want to know what is the correct answer and why it's true ? (Please answer in simple words)
Please Help!
 A: There are many ways to assign a finite number to infinite series. Taking their actual sum (defined by taking the limit of the partial sums) is one, but there are other (much more theoretical ways) than just sums. Many of them give answers to series that are not actually convergent. The best of them agree with the sums if the series is convergent and give finite answers on some series that are not convergent. See for instance Cesaro summation and Abel summation.
Most of the methods that agree with sums on convergent series but give a finite answer to $\sum_{n = 0}^\infty (-1)^n$ do indeed give the answer $\frac12$. But those are not sums in the strict sense.
By the way, if you ever happen to come across the statement $\sum_{n = 1}^\infty n = -\frac1{12}$, my answer will be the same, and I've given it several times on that exact question on this site.
A: Correctness of answer depends on meaning the sum of an infinite series. According to wiki, nowadays in math the sum of an infinite series is the limit of of the sequence of its partial sums. In this case we see that Grandi's series is divergent.
https://en.wikipedia.org/wiki/Grandi%27s_series
A: From Wikipedia:  
We know the sum of an infinite series is defined to be the limit of the sequence of its partial sums, if it exists. The sequence of partial sums of Grandi's series is $1, 0, 1, 0, ...$, which clearly does not approach any number (although it does have two accumulation points at $0$ and $1$). Therefore, Grandi's series is divergent.  

Cesaro's method of summing up divergent series: The basic idea is similar to Leibniz's probabilistic approach: essentially, the Cesàro sum of a series is the average of all of its partial sums. Formally one computes, for each $n$, the average $\sigma_{n}$ of the first $n$ partial sums, and takes the limit of these Cesàro means as $n$ goes to infinity.
For Grandi's series, the sequence of arithmetic means is
$1, 1/2, 2/3, 2/4, 3/5, 3/6, 4/7, 4/8, …$
or, more suggestively, $(1/2+1/2), 1/2, (1/2+1/6), 1/2, (1/2+1/10), 1/2, (1/2+1/14), 1/2, …$
where $\sigma _{n}={\frac {1}{2}}$ for even $n$ and $\sigma _{n}={\frac {1}{2}}+{\frac {1}{2n}}$ for odd $n$.  
This sequence of arithmetic means converges to $1/2$, so the Cesàro sum of $\sum a_{k}$ is $\frac{1}{2}$. Equivalently, one says that the Cesàro limit of the sequence $1, 0, 1, 0, …$ is $\frac{1}{2}$.
A: The series is divergent. It would converge if and only if there were some real number $l$ with the property that $l=\mathrm{lim}_{k\rightarrow\infty}\Sigma_{n=0}^{k} (-1)^{n}$. If that were the case then for every real number $\epsilon>0$, however small, we could find a non-negative integer $N$ with the property that $\vert\Sigma_{n=0}^{k} (-1)^{n}-l\vert<\epsilon$ for all $k$ such that $k\geq N$. In particular it would then follow that if $k_{1}$ and $k_{2}$ are any two non-negative integers such that $k_{1}, k_{2} \geq N$, we would have $\vert \Sigma_{n=0}^{k_{1}} (-1)^{n} - \Sigma_{n=0}^{k_{2}} (-1)^{n} \vert<2\epsilon$. But then consider the case where $\epsilon=\frac{1}{2}$. Then it would follow that for some sufficiently large non-negative integer $N$ we can never have two non-negative integers $k_{1}, k_{2}$ such that $k_{1}, k_{2} \geq N$ and $\vert \Sigma_{n=0}^{k_{1}} (-1)^{n} - \Sigma_{n=0}^{k_{2}} (-1)^{n} \vert =1$. But clearly it is always possible to find two non-negative integers $k_{1}, k_{2}$ which will make that happen no matter how large $N$ is. So we must conclude that the limit of the partial sums of the series does not exist and the series is not convergent. This means that according to standard usage the expression $\Sigma_{n=0}^{\infty} (-1)^{n}$ is not well-defined.
A: It's just a matter of definitions. If you define the symbol $\sum_{n=0}^\infty (-1)^n$ as the limit of its partial sums (e.g. the limit of the sequence $(1), (1 -1), (1-1+1), (1-1+1-1), \cdots)$, which is the usual definition, then the sum is divergent. But if you define the symbol as $\sum_{n=0}^\infty (-1)^n$ something else (such as Cesaro Summation definition), then you can get this symbol being equal to other values (including $\dfrac 12$)
