Using the squareroot on small o (landau notation)

Question

Suppose $k\rightarrow\infty$, and $\sqrt(\frac{1}{x_k})=o(y_k)$, is it then true that $\frac{1}{x_k} =o(y_k)$?

$x_k\rightarrow\infty$ and $y_k\rightarrow 0$, if it helps.

Help would be appreciated.

Answer 1

Recall the definition of little-o notation: we say $f(x) \in o(g(x))$ if $$ \frac{f(x)}{g(x)} \rightarrow 0 \text{ as } x \rightarrow \infty .$$ You are given that $\sqrt{\frac{1}{x_k}} \in o(y_k)$, meaning that $$ \frac{\sqrt{\frac{1}{x_k}}}{y_k} \rightarrow 0 \text{ as } k \rightarrow 0 .$$ From this you can see $$ \frac{\frac{1}{x_k}}{y_k} \leq \frac{\sqrt{\frac{1}{x_k}}}{y_k} \rightarrow 0 \text{ as } k \rightarrow 0 $$ since $x_k \geq 1$ for sufficiently large $k$, and we know that for any $0 < \alpha < 1$, $\alpha < \sqrt{\alpha}$.

That's the mathematical proof; now what's the mathematical reasoning? Well, $f(x) \in o(g(x))$ means that $f$ grows much slower than $g$. We are given that $\sqrt{\frac{1}{x_k}}$ grows much slower (in $k$) than $y_k$. Since $\frac{1}{x_k}$ grows slower than $\sqrt{\frac{1}{x_k}}$ (by the previous observation that for any $0 < \alpha < 1$, $\alpha < \sqrt{\alpha}$), we know that $\frac{1}{x_k}$ grows slower than $y_k$.

In fact we know something stronger: $\frac{1}{x_k} \in o(\sqrt{\frac{1}{x_k}})$. Use this together with $$ f(x) \in o(g(x)) \text{ and } g(x) \in o(h(x)) \text{ implies } f(x) \in o(h(x))$$ to get the desired result.