Prove by induction $\sum \frac {1}{2^n} < 1$

Prove by induction $\sum \frac {1}{2^n} < 1$

Well supposing the base case has been shown to be true, I start with the induction step:

Suppose true for n = k:

$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} < 1$$

Want to show this is true for: $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} < 1$$

Now this is where i am getting stuck, should i be tryig to show that $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} < 1 +\frac{1}{2^{k+1}}$$ or should i be attempting to show that $1 + \frac{1}{2^{k+1}} < 1$ which is utter nonsense. So i am stuck on what exactly to prove here.

Note i also thought of maybe using the geometric series of $\frac {1}{2} $ as some upper bound but then that would provide me a value that was greater than 1 so it didn't work out

Solution 1:

For $n=k+1$:

$$\frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k}\right) < \frac{1}{2} + \frac{1}{2}\times 1 =1.$$

Solution 2:

Use induction to prove that $$\frac 12 + \frac 14 + \cdots + \frac 1{2^n} = 1 - \frac{1}{2^n}$$ for all $n$. This will give you the result immediately since $1 - \frac{1}{2^n} < 1$.