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Can we say that $k$ grows faster than $\sqrt{k}$ when term is large? But what is the formal way write it ?

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Note that, for all positive integer $k$, $$\frac{1}{k-2\sqrt{k}}=\frac{\frac{1}{k}}{1-2\frac{\sqrt{k}}{k}}=\frac{\frac{1}{k}}{1-2\frac{1}{\sqrt{k}}}$$

Thus, if you know that $$\lim_{k\to\infty}\frac{1}{k}=0\quad\text{and}\quad\lim_{k\to\infty}\frac{1}{\sqrt{k}}=0,$$ then you can conclude that $$\lim_{k\to\infty}\frac{1}{k-2\sqrt{k}}=\frac{0}{1-2\cdot 0}=0$$

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You can factor a $\sqrt{k}$ out of the bottom to get: $\lim_{k \to \infty} \frac{1}{\sqrt{k}(\sqrt{k}-2)}$. Now it should be clear that the bottom goes to infinity.

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We can restrict $k > 4$, then it suffices to show that $$ \lim_{k \to \infty} (k - 2\sqrt{k}) = \infty $$ Using the fact that $2\sqrt{k} \in o(k)$, $$ \lim_{k \to \infty} (k - 2\sqrt{k}) = \lim_{k \to \infty} k = \infty $$

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  • $\begingroup$ But how do you know $ \lim_{k \to \infty} (k - 2\sqrt{k}) = \infty$? $\endgroup$ – CoolKid Mar 30 '16 at 0:09
  • $\begingroup$ @CoolKid I have shown it in the second step. $\endgroup$ – Henricus V. Mar 30 '16 at 0:09

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