show by induction that, $1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}=2-\frac{1}{2^n}$ for positive integers n.
for the base case, I am doing, if $n=1$; $LHS=\frac{1}{2^1}= \frac{1}{2}$
$RHS= 2- \frac{1}{2}= 2-\frac{1}{2}= \frac{3}{2}$
which is not the same. What am I doing wrong?
Think of the sequence as the following sum:
$$\sum_{i=0}^n 2^{-i}$$
So with that:
If $n=0$: $\sum_{i=0}^n 2^{-i} = 2^{-0}=2^0=1$ .. which indeed equals $2-\frac{1}{2^0}=2-\frac{1}{1}=2-1=1$
(so, interestingly, the claim even holds for $04, and not just the positive integers)
For $n=1$ we get:
$\sum_{i=0}^n 2^{-i} = 2^{-0}+2^{-1} = 1+\frac{1}{2}$ ... and that indeed equals what you found for the RHS.
Moral: Always try to state the claim that you are trying to prove as precisely as possible!
