How to prove that $$f = \frac{2^{tK} -1}{K}$$ is increasing with $K$, where $t > 0$ and $K = 1, 2, 3, \dotsc$? That is, I want to prove that the above expression is maximized as $K \to \infty$.
My attempt:
Relaxing $K > 0$ to be continuous, I tried showing that the derivative $$\frac{\partial f}{\partial K} = \frac{(\ln(2)tK-1)2^{tK}+1}{K^2} > 0.$$ This means that we have to show that: $$(\ln(2)tK-1)2^{tK}+1 > 0.$$ But I have been unable to show this. What I tried further unsuccessfully is that, to show $(\ln(2)tK-1)2^{tK}+1 > 0$ as $t \to 0_{+}$, because if the inequality holds for $t \to 0_{+}$, it holds for any $t > 0$.
Let $s=tln(2)$. Then $f=\frac{e^{sK}-1}{K}=\frac{1}{K}(\sum_{j=1}^\infty \frac{(sK)^j}{j!})$ Each term in the sum increases with $s$. Therefore $f$ increases with $t$ for each $K$. Also for fixed $s$, each term from $j=2$ onward, after dividing by $K$, increases with $K$.