sequence question

**rpatel** · April 12th 2010, 06:48 AM

hi i didn't understand the following question

Let $\varepsilon>0$ be given. For the following sequence $(a_{n})_{n\in\mathbb{N}}$ find a natural number $N_{\varepsilon}$ such that $\forall\geq N_{\varepsilon}$ , $\vert a_{n}\vert<\varepsilon$ thereby showing that $a_{n}\rightarrow0$ as $n\rightarrow\infty$ .

a) $a_{n}=\frac{n+\sqrt{n}}{n^{2}+1}$

thanks

**Failure** · April 12th 2010, 07:10 AM

Originally Posted by rpatel

hi i didn't understand the following question

Let $\varepsilon>0$ be given. For the following sequence $(a_{n})_{n\in\mathbb{N}}$ find a natural number $N_{\varepsilon}$ such that $\forall {\color{red}n}\geq N_{\varepsilon}$ , $\vert a_{n}\vert<\varepsilon$ thereby showing that $a_{n}\rightarrow0$ as $n\rightarrow\infty$ .

a) $a_{n}=\frac{n+\sqrt{n}}{n^{2}+1}$

thanks

Consider:
$|a_n|=\frac{n+\sqrt{n}}{n^2+1}=\frac{1+\sqrt{\frac {1}{n}}}{n+\frac{1}{n}} \leq \frac{2}{n+\frac{1}{n}}< \frac{2}{n}< \varepsilon$
Overall we get $|a_n|<\frac{2}{n}<\varepsilon$
Solving the last inequality for $n$ gives $\frac{2}{\varepsilon}<n$ . Therefore you can choose $N_\varepsilon := \left\lceil \frac{2}{\varepsilon}\right\rceil$ , or some bigger number, of course.

**rpatel** · April 12th 2010, 07:16 AM

Hi i was wondering if you could please explain the step from
$\frac{n+\sqrt{n}}{n^2+1}=\frac{1+\sqrt{\frac{1}{n} }}{n+\frac{1}{n}}$

and from $\frac{1+\sqrt{\frac{1}{n}}}{n+\frac{1}{n}} \leq \frac{2}{n+\frac{1}{n}}$ in this case where does the 2 come from i undestand that is true statement but why number 2 on the top of the fraction and not 3 for example.

thanks

**Deadstar** · April 12th 2010, 07:31 AM

Originally Posted by rpatel

Hi i was wondering if you could please explain the step from
$\frac{n+\sqrt{n}}{n^2+1}=\frac{1+\sqrt{\frac{1}{n} }}{n+\frac{1}{n}}$

and from $\frac{1+\sqrt{\frac{1}{n}}}{n+\frac{1}{n}} \leq \frac{2}{n+\frac{1}{n}}$ in this case where does the 2 come from i undestand that is true statement but why number 2 on the top of the fraction and not 3 for example.

thanks

Since Failure is offline I'll post up an answer.

For the first question you had, he multiplied the fraction by $\frac{1/n}{1/n}$ . I.e, divided everything by n.

for the second question, notice that the maximum value of $\sqrt{\frac{1}{n}}$ for $n \in \mathbb{N}$ is 1. Hence,

$1 + \sqrt{\frac{1}{n}} \leq 1+1 = 2$ .

**rpatel** · April 12th 2010, 07:52 AM

oh ok yep i understand thanks.

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