Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.
We start with $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ for all positive integers.
I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.
This is equivalent to $\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$
Let $P(n)$ be summation shown above.
Base case for $n=1$, the first positive integer,
$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$
so base case is true.
Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus
$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$
This can be written as
$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$
I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes.