<s>[INST] Are there contradictions in math? Someone told me that math has a lot of contradictions. 
He said that a lot of things are not well defined.
He told me two things that I do not know.


*

*$1+2+3+4+...=-1/12$

*what is infinity $\infty$?


Since I am not a math specialist and little. How to disprove the previous two?
How can I convince him?
 [/INST] There is already great general answer by Ittay Weiss, so I will try a different approach. In fact, I will try to explain a bit the infinite sum you stated. As for the infinity, one could write a lot about it (mainly because there are multiple infinities, each with different properties), and I couldn't even hope to fit it here. Unfortunately, I don't know any sources or text about infinity that I would like to recommend.

Consider the following infinite sum
$$G(x) = x^0 + x^1 + x^2 + x^3 + \ldots = \sum_{i = 0}^{\infty}x^i.$$
We have no idea what such a sum might mean (without any proper definition, which wasn't stated). However, suppose that we could manipulate it in a way similar to numbers. Let's start with multiplying by $x$
$$x\cdot G(x) = x^1 + x^2 + x^3 + x^4 + \ldots = \sum_{i = 0}^{\infty}x\cdot x^{i},$$
and then add $x^0 = 1$ (we assume that $x \neq 0$)
$$1 + x\cdot G(x) = x^0 + x^1 + x^2 + x^3 + x^4 + \ldots = x^0 + \sum_{i = 1}^{\infty}x^{i} = \sum_{i = 0}x^i.$$
We got the same expression! In other words, we don't know what $G(x)$ means yet, but if it does mean anything similar to a number, then this number has to satisfy
$$G(x) = 1+x\cdot G(x),\quad \text{ that is, }\quad G(x) = \frac{1}{1-x}.$$
Indeed, there is a way to specify what $G(x)$ means for $|x| < 1$ and then, e.g. $$G\left(\frac{1}{2}\right) = 1+\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = \frac{1}{1-\frac{1}{2}} = 2.$$
On the other hand, for other values of $x$ we get peculiar results, like
$$G(2) = 1 + 2 + 4 + 8 + \ldots = \frac{1}{1-2} = -1,$$
which would mean that sum of positive numbers got negative. However, this is because we didn't defined what $G(2)$ would mean, we only said we could do it for $|x| < 1$. Nevertheless, there is a way of dealing with this (e.g. see $p$-adic numbers), but for $x=1$ we get even stranger result, that is,
$$G(1) = 1 + 1 + 1 + \ldots = \frac{1}{1-1} = \frac{1}{0} =\ ?!$$
Similar situation happens with your sequence. Consider
\begin{align}
S(x) &= 0\cdot x^0 + 1\cdot x^1 + 2\cdot x^2 + 3\cdot x^3 + \ldots = \sum_{i = 0}^{\infty} i \cdot x^i \\
x\cdot S(x) &= \color{white}{0 \cdot x^{?}} + 0 \cdot x^1 + 1 \cdot x^2 + 2 \cdot x^3 + \ldots = \sum_{i=0}^{\infty} i\cdot x^{i+1} \\
x\cdot G(x) + x\cdot S(x) &= 0\cdot x^0 + 1\cdot x^1 + 2\cdot x^2 + 3\cdot x^3 + \ldots = \sum_{i = 0}^{\infty} i \cdot x^i \\
S(x) &= x\cdot G(x) + x \cdot S(x) \\
S(x) &= \frac{x \cdot G(x)}{1-x} = \frac{x}{(1-x)^2}
\end{align}
which is true for $|x| < 1$ (i.e. we can define the infinite sum in a way that would make the above equality true with an appropriate $x$), but with $x = 1$ we get
$$ S(1) = \frac{1}{(1-1)^2} = \frac{1}{0} = \ ?! $$
On the other hand, observe that
\begin{align}
 S(1) &= 1 + 2 + 3 + 4 + 5 + 6 + \ldots \\
4S(1) &= \color{white}{0} + 4 + \color{white}{0} + 8 + \color{white}{0} + 12 + \ldots \\
-3S(1) &= 1 - 2 + 3 - 4 + 5 - 6 + \ldots \\
-S(-1) &= 1 - 2 + 3 - 4 + 5 - 6 + \ldots = -\frac{-1}{\big(1-(-1)\big)^2} = \frac{1}{4} \\
-3S(1) &= -S(-1) = \frac{1}{4} \\
S(1) &= -\frac{1}{12}
\end{align}
which means that, if there is any number-like interpretation of $S(1)$ (and $S(-1)$ which we have used), then it has to be equal $-\frac{1}{12}$. However, note that we didn't argue whether there actually is such an interpretation. Nevertheless, those interpretations of $S(x)$ which do allow $S(1)$ also explain why the sum of positive values might be considered negative (i.e. given particular interpretation it stops being strange). If you would like to know more about it, you can start here.
I hope this helps $\ddot\smile$
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