Adding always the half of the previous number to a number beginning with 1... This question blows my mind.
Let's say we start at 1 and keep adding always 0.5 * the previous number.
1 + 0.5 + 0.25 + 0.125 + ..... 
The question is: will it reach infinity?
I really don't know, kept thinking about this alot.

Consider calculating the sum in steps, at each step adding the next term. At the $n^{th}$ step, you're adding $1/2^n$ (adding $1$ is the $0^{th}$ step). Then at the $n^{th}$ step, the total sum is the following:
$$
s_n = 1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n}
$$
Now remark that you can perform the following calculation:
\begin{align*}
1 + \frac{1}{2} s_n &= 1 + \frac{1}{2}
\left(
1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n}
\right)
\\
&=
1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n} + \frac{1}{2^{n+1}}
\\
&= s_n + \frac{1}{2^{n+1}}
\end{align*}
Then if you solve for $s_n$ using elementary algebra (which you can do since $s_n$ is finite), you end up with:
$$
s_n = 2 - \frac{1}{2^n} < 2
$$
What we have just shown is that $s_n \leq 2$ for any $n$ that we choose. In other words, no matter how many terms we add, the resulting sum will always be less than $2$. So, the sum will not reach infinity.
