Inequality

Let $b_{k}$ = $3^k$ $a_{k}$ , then Cauchy-Schwarz says $∑_{k=1}^n{a_{k}2^k} = ∑_{k=1}^n$ $b_{k}$ $({\frac{2}{3}})^k$

$<=({∑_{k=1}^n{b_{k}}^2})^{1/2}({∑_{k=1}^n({\frac{4}{9}})^k})^{1/2}$

$=({∑_{k=1}^n{{{a_{k}}^2}9^k})^{1/2}}{\frac{2}{\sqrt{5}}}{\sqrt{1-({\frac{4}{9}})^n}}$

$<={2{\sqrt{1-({\frac{4}{9}})^n}}}$

In the last expression, as n tends to infinity, $\frac{4}{9}$ being a positive fraction, it's nth power goes to 0, so that the limit of the best positive constant is 2.

Here the last expression is the best positive constant because by Cauchy Schwartz and the given condition we have a condition where equality actually occurs.