How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}} = \frac{2^{n+1}-1}{2^{n+1}}$.
What should the approach be here?

Just in case you may be interested. More generally you can get a similar expression for such sums of powers of inverses. They basically come from the factorization
$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \cdots + a + 1)$$
which can be proved by multiplying out the right hand side and then most of the terms cancel out and you'll be left with the expression on the left.
Then from this factorization, by dividing by $a - 1$ you get
$$\frac{a^n - 1}{a - 1} = a^{n-1} + a^{n-2} + \cdots + a + 1$$
Now by dividing by $a^n$ you get
$$
\frac{a^n - 1}{a^{n}(a - 1)} = \frac{a^{n-1}}{a^n} + \frac{a^{n-2}}{a^n} + \cdots + \frac{a}{a^n} + \frac{1}{a^n}
$$
$$
= \frac{1}{a} + \frac{1}{a^2} + \cdots \frac{1}{a^{n-1}} + \frac{1}{a^n}
$$
So that you have
$$
\boxed{
\displaystyle\frac{a^n - 1}{a^{n}(a - 1)} = \frac{1}{a} + \frac{1}{a^2} + \cdots \frac{1}{a^{n-1}} + \frac{1}{a^n}
}
$$
Therefore the expression you want to prove by induction is obtained just by setting $a = 2$, but now you can give different values to $a$ and get similar expressions.
