Adding always the half of the previous number to a number beginning with 1...

Question

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.

Answer 1

So what you're looking for is the value of this series, which will call it $S$.

$$S=\sum_{n=0}^{\infty} \left( \frac{1}{2} \right)^n=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$$

If we divide this value $S$ by $2$ we get

$$\frac{S}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=S-1 \Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $\infty$.

Answer 2

Rewrite it with fractions and it might be a bit clearer:

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} ...$

Now calculate the partial sums:

$1$, $1 \frac{1}{2}$, $1 \frac{3}{4}$, $1 \frac{7}{8}$, $1 \frac{15}{16}$, ...

See a pattern? Can you see this going over $2$? How close do you think it might get?

Answer 3

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.