Inductively prove $1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$. Inductively prove that the formula holds for all $n\in\Bbb{N}$:
$$1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.$$
What I have so far:
base: n = 1: $$1+\frac{1}{2}=2-\frac{1}{2}=1.5$$
inductionstep: n = k: $$1+\frac{1}{2}+\cdots+\frac{1}{2^k}=2-\frac{1}{2^k}$$
inductionhypothesis: n=k+1: $$1+\frac{1}{2}+\cdots+\frac{1}{2^k+1}=1+\frac{1}{2}+\cdots+\frac{1}{2^k}+\frac{1}{2^(k+1)}=(2-\frac{1}{2})+\frac{1}{2^k+1}$$
This is where I am stuck and not sure what to do next.
 A: The formula
$$
1+\dots+\frac{1}{2^n}=2-\frac{1}{2^n}
$$
is true for $n=0$ (in which case the LHS of the equation actually only has one term in it). Therefore, it is better to choose $n=0$ as your base case. Now suppose it is true when $n=k$, that is
$$
1+\dots+\frac{1}{2^k}=2-\frac{1}{2^k} \, .
$$
Then, for $n=k+1$, we have
\begin{align}
1+\dots+\frac{1}{2^{k+1}}&=\left(1+\dots+\frac{1}{2^k}\right)+\frac{1}{2^{k+1}} \\[5pt]
&= 2-\frac{1}{2^k}+\frac{1}{2^{k+1}} \\[5pt]
&= 2-\frac{2}{2^{k+1}}+\frac{1}{2^{k+1}} \\[5pt]
&= 2-\frac{1}{2^{k+1}} \, .
\end{align}
Hence, if the statement is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=0$, by the principle of mathematical induction it must be true for all nonnegative integers $n$.
A: Use the following structure:
Step 1. $n = 0$, means $1 = 2 - \frac{1}{2^0} = 2 - 1 = 1$, which is true. We can proceed to induction step.
Step 2. Use simple induction: $p(n) \to p(n + 1)$. So, if:
$$1 + \dots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$$
We have:
$$1 + \dots + \frac{1}{2^n} + \frac{1}{2^{n + 1}} = 2 - \frac{1}{2^n} + \frac{1}{2^{n+1}} = 2 - \frac{2}{2^{n + 1}} + \frac{1}{2^{n + 1}} = 2 - \frac{1}{2^{n + 1}}$$
This ends the proof. $\blacksquare$
