Why this :$\sum\limits_{n=0}^{\infty}(0)=0$? I'm confused how :$\sum\limits_{n=0}^{\infty} 0=0$ but after evaluation we w'd get $0.\infty$ which it's indeterminate case and wolfram alpha show here that is a convergent series which it's equal's $0 $, i'm only sure  that is $0$ for it partial sum .
My question here is : Is there somone who show me how do I convince my self by :$\sum\limits_{n=0}^{\infty} 0=0$ ?
Thank you for any help 
 A: Consider the sequence $x_i = 0$ for each $i$.
Moreover, consider the sequence:
$$s_n = \sum_{i=0}^n x_i.$$
It's clear that $s_n = 0$ for each $n$. This is "clear" since the sum $s_n$ is finite. Indeed, $s_n$ is the sum of $n+1$ zeros, no?
Now, what about:
$$\lim_{n \to +\infty} s_n?$$
For sure it is $0$, since $s_n = 0$ for all $n$!!!
Don't forget that:
$$\lim_{n \to +\infty} s_n = \lim_{n \to +\infty}\sum_{i=0}^n x_i = \sum_{i=0}^{+\infty} x_i = \sum_{i=0}^{+\infty} 0.$$
A: The thing is that $\sum_{n=0}^\infty a_n$ is just a "formal object".  We might give it some value if some limit exists (for example the limit of partial sums), but we should distinguish between the formal object and the limit.  You could come up with your own versions of what numbers to assign series.  You should search on Youtube for example for the "proof" that $\sum_{n=0}^\infty n = \frac{-1}{12}$, it is in fact entirely reasonable to assign that value, but it is not a limit of partial sums.  The standard is to take a limit of partial sums.  That has flaws, but that's the definition.  And the limit of zeros is zero.
So the thing is to never think that a series is a sum of infinitely many numbers.  We can't sum infinitely many numbers, we can only sum finitely many.  All we do is define some way of getting something that sort of kind of behaves as we think a sum of infinitely many numbers should behave, but you cannot generalize everything you know about finite sums to this new operation.
A: By definition, one has
$$
\sum_{n=0}^\infty a_n = \lim_{N \to \infty}\sum_{n=0}^N a_n 
$$ giving
$$
\sum_{n=0}^\infty 0 = \lim_{N \to \infty}\sum_{n=0}^N 0= \lim_{N \to \infty}0=0.
$$
A: Note that $\sum_{k=0}^\infty a_k$ is defined as a limit,
$$\sum_{k=0}^\infty a_k = \lim_{n\to\infty} \sum_{k=0}^n a_k$$
Now in a limit, if inserting gives you an indefinite value, this does not mean the limit is necessarily undefined, but that you have to do something else to find whether a limit exists and what it is.
Another example of an indefinite form coming out of a limit, where the limit clearly exists:
$$\lim_{x\to 1}\frac{x^2-1}{x-1}=\lim_{x\to 1}(x+1) = 2$$
But when you just insert $x=1$ into the expression, you get
$$\frac{x^2-1}{x-1} = \frac{1-1}{1-1} = \frac{0}{0}$$
which is an indefinite form.
So after seeing the indeterminate form, the correct reaction is not "that cannot be defined" but "I must do the limit explicitly". And doing so for the infinite sum, you'll find
$$\sum_{k=0}^\infty 0 = \lim_{n\to\infty}\sum_{k=0}^n 0 = \lim_{n\to\infty} 0n = \lim_{n\to\infty} 0 = 0$$
Note that obtaining an indeterminate form when trying to calculate the limit by insertion doesn't guarantee a limit exists; for example in $\lim_{x\to 0} x/x^2$ inserting $x=0$ gives an indeterminate form, and the limit doesn't exist. All an indeterminate form tells you is: Try something different.
