by Guest » Wed Apr 17, 2019 8:59 am
"Induction" is pretty standardized! You want to use "proof by induction" to prove that [tex]\frac{n}{2^n}< 2[/tex] for n any positive integer. When n= 1 that says [tex]\frac{1}{2^1}= \frac{1}{2}< 2[/tex] which is certainly true.
Now, assume that, for some positive integer, k, [tex]\frac{k}{2^k}< 2[/tex]. We want to use that to prove that [tex]\frac{k+1}{2^{k+1}}< 2[/tex]. First an obvious step- from [tex]\frac{k}{2^k}< 2[/tex], [tex]\frac{k+1}{2^k}= \frac{k}{2^k}+ \frac{1}{2^k}< 3[/tex] since [tex]\frac{1}{2^k}< 1[/tex]. Now, divide both sides of [tex]\frac{k+1}{2^k}= \frac{k}{2^k}+ \frac{1}{2^k}< 3[/tex] by 2 to get [tex]\frac{k+1}{2^{k+1}}< \frac{3}{2}< 2[/tex].