Proof that $\displaystyle \lim_{i \to \infty} \bigg | \frac{1}{\sqrt{i}}\bigg |=0 $

Let $\epsilon>0$ be arbitrary and let $$N= \begin{cases} 1 & \text{ if } 1 \geq \epsilon\\ \epsilon & \text{otherwise.} \end{cases}$$

Note: whether or not $1 \geq \epsilon$, it follows that $N \geq 1$ and $N \geq \epsilon$.

Assume $i>N$

Since $i>N\geq 1$, $\sqrt{i}>\sqrt{N}\geq \sqrt{1}$.

We then have $\bigg |\frac{1}{\sqrt{i}}\bigg |=\frac{1}{|\sqrt{i}|}=\frac{1}{\sqrt{i}}<1 \leq N$

My main worry about this proof is near $\frac{1}{\sqrt{i}}<1$. My justification here is that $i>N \geq 1$ and for any number $t>1$ we get $\sqrt{t}>\sqrt{1}=1$. Furthermore, for numbers $a$ and $b$, if $a>b>0$ then $\frac{b}{a} < 1$. Thus $\sqrt{i}>1>0$ so $\frac{1}{\sqrt{i}} <1$.

  • $\begingroup$ When calculating a limit, note that if $N$ fits to some $\varepsilon$ then it necessarily fits any $\varepsilon'>\varepsilon$. $\endgroup$ – Asaf Karagila May 9 '11 at 6:43
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    $\begingroup$ You seem to prove that $\bigg |\frac{1}{\sqrt{i}}\bigg |\leq N$, but you need to prove it $\leq \epsilon$. $\endgroup$ – ShreevatsaR May 9 '11 at 6:44
  • $\begingroup$ can we not just choose N = int(1/E^2) + 1 or something simple and show it in one line? $\endgroup$ – The Chaz 2.0 May 9 '11 at 6:46
  • $\begingroup$ Yes we can. $ $ $\endgroup$ – Did May 9 '11 at 7:45

As mentioned in the comments, you are confused about the definition of a limit. Suppose we have a sequence $\{u_n\}$, and want to prove that $u_n \rightarrow L$. You need to show that for any $\epsilon > 0$, there is a number $N$ such that for all natural numbers $n > N$ $$ |u_n-L| < \epsilon. $$ The number $L$ is called the limit.

In this question, that is the same as showing $$ \frac{1}{\sqrt{n}} < \epsilon. $$

To prove this, suppose $\epsilon>0$ and then let $n > N = \epsilon^{-2}$. (Note that you can't use $N$ as an index, it is not in general a natural number. This may differ with local usages.) Then $$ \frac{1}{\sqrt{n}} < \frac{1}{\sqrt{\epsilon^{-2}}} = \epsilon, $$ as required.


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