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My question is

If there are positive numbers $a<1$ and $\infty>C>0: f(i)\leq Ca^i$, does it holds that there are finite positive numbers $C_0,b:f(i)\leq C_0i^{-b}$, where it is assumed that $f\geq 0$, bounded and nonincreasing. Equivalently,

$$\exists C>0,a<1:f(i)\leq Ca^i\implies \exists C_0,b>0:f(i)\leq C_0i^{-b},$$ for any positive $i$, where $f$ is a real nonnegative, bounded and nonincreasing function?

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1 Answer 1

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Observe that for every $i \geq 0$ one has $a^i \leq \frac{C_1}{i}$, where $C_1 > 0$ is un ugly constant depending on $a$. Indeed, the function $ia^i$ is 0 in 0, tends to 0 at $+\infty$ and by differentiating we find its maximum in the point $i_0$ such that $$ a^{i_0} + i_0 \log a a^{i_0} = 0 \implies i_0 = \frac{-1}{\log a}, $$ where the value of the function is $\frac{-a^{\frac{-1}{\log a}}}{\log a} = C_1 > 0$. With this definition you can take $C_0 = CC_1$ and $b = 1$ to get $f(i) \leq Ca^i \leq CC_0 i^{-1}$.

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