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Thread: sequence question

  1. #1
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    sequence question

    hi i didn't understand the following question

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

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

    thanks
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by rpatel View Post
    hi i didn't understand the following question

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

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

    thanks
    Consider:
    $\displaystyle |a_n|=\frac{n+\sqrt{n}}{n^2+1}=\frac{1+\sqrt{\frac {1}{n}}}{n+\frac{1}{n}} \leq \frac{2}{n+\frac{1}{n}}< \frac{2}{n}< \varepsilon$
    Overall we get $\displaystyle |a_n|<\frac{2}{n}<\varepsilon$
    Solving the last inequality for $\displaystyle n$ gives $\displaystyle \frac{2}{\varepsilon}<n$. Therefore you can choose $\displaystyle N_\varepsilon := \left\lceil \frac{2}{\varepsilon}\right\rceil$, or some bigger number, of course.
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  3. #3
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    Hi i was wondering if you could please explain the step from
    $\displaystyle \frac{n+\sqrt{n}}{n^2+1}=\frac{1+\sqrt{\frac{1}{n} }}{n+\frac{1}{n}}$

    and from$\displaystyle \frac{1+\sqrt{\frac{1}{n}}}{n+\frac{1}{n}} \leq \frac{2}{n+\frac{1}{n}}$ 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
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by rpatel View Post
    Hi i was wondering if you could please explain the step from
    $\displaystyle \frac{n+\sqrt{n}}{n^2+1}=\frac{1+\sqrt{\frac{1}{n} }}{n+\frac{1}{n}}$

    and from$\displaystyle \frac{1+\sqrt{\frac{1}{n}}}{n+\frac{1}{n}} \leq \frac{2}{n+\frac{1}{n}}$ 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 $\displaystyle \frac{1/n}{1/n}$. I.e, divided everything by n.

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

    $\displaystyle 1 + \sqrt{\frac{1}{n}} \leq 1+1 = 2$.
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  5. #5
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    oh ok yep i understand thanks.
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