<s>[INST] Sum of $1+2+4+8+....$ I was solving a recurrence problem which had a sequence such as $y = (1+2+4+8+...)\sqrt n$, and I wanted to find what $x = 1+2+4+8+...$ was. So consider $x = 1+2+4+8+...$ as an infinite series.
$$x-1 = 2+4+8+...$$
$$2(x)=2(1+2+4+8+...) = 2+4+8+... = x-1$$
$$2x = x-1 \Rightarrow x=-1$$
What does $x=-1$ represent? Have I made a mistake in my calculations? Of course, we expect that $x=\infty$ since the sum grows by a factor of $2$ with each term.
 [/INST] Here you manipulate x as if it is finite numbers, so you have $2x$. But in fact $n\cdot\infty = \infty$ for any finite number $n$. And also $n+\infty=\infty$ again for any finite number $n$. So basically your equation is $\infty=\infty$, which have nothing useful.
You might be interested in this proof which shows that:
$1+2+3+...=-1/12$
The proof set $c=1+2+3+...$ and have $4c=4+8+12+...$, then he got $c-4c = 1-2+3-4+...=1/(1+1)^2$, ans so $c=-1/4$, which is absurd.
As said in wikipedia I linked to, 

It is dangerous to manipulate infinite series as if they were finite sums.

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