<s>[INST] 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.
 [/INST] 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. $\blacksquare$
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