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Thread: Existence of limit

  1. #1
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    Existence of limit

    Hello all,

    I have the following problem: suppose that $\displaystyle \{S_{n} \}$ is a convergent sequence. Then we are asked to prove that $\displaystyle \lim_{n\rightarrow \infty}2S_{n}$ exists.

    I have done the following:

    By assumption I know that $\displaystyle \{S_{n} \}$ converges. This means that $\displaystyle \forall \epsilon>0$ we can find an $\displaystyle N\in \mathbb{N}$ such that $\displaystyle \mid S_{n}-L \mid<\epsilon$ whenever $\displaystyle n\geq N$.
    We therefore have:
    $\displaystyle \mid S_{n}-L \mid<\epsilon$
    $\displaystyle 2\cdot\mid S_{n}-L \mid<2\epsilon$
    $\displaystyle \mid 2S_{n}-2L \mid<2\epsilon$
    $\displaystyle \mid 2S_{n}-2L \mid-\epsilon<2\epsilon-\epsilon$
    $\displaystyle \mid 2S_{n}-2L \mid-\epsilon<\epsilon$

    Could someone point in the right direction?
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  2. #2
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    Re: Existence of limit

    Quote Originally Posted by detalosi View Post
    I have the following problem: suppose that $\displaystyle \{S_{n} \}$ is a convergent sequence. Then we are asked to prove that $\displaystyle \lim_{n\rightarrow \infty}2S_{n}$ exists.
    Could someone point in the right direction?
    We are given that $(s_n)\to L$
    If $\varepsilon>0 $ then $\dfrac{\varepsilon}{2}>0$ so $\exists N\in \mathbb{Z}^+$ such that $n > N \Rightarrow |{s_n} - L| < \dfrac{\varepsilon}{2} $
    $|2{s_n} - 2L|=|s_n-L|(2) <(2) \dfrac{\varepsilon}{2}=\varepsilon$ By definition $(2s_n)\to 2L$
    Thanks from topsquark and detalosi
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  3. #3
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    Re: Existence of limit

    Thanks.

    I have written it in the following form:

    Since $\displaystyle \{ S_{n} \}$ converges to $\displaystyle L$ then $\displaystyle \forall \epsilon >0$ there is an $\displaystyle N\in \mathbb{N}$ such that $\displaystyle \mid S_{n}-L \mid <\epsilon$ whenever $\displaystyle n\geq N$. Since this is true for all $\displaystyle \epsilon>0$ then it is also true for $\displaystyle \frac{\epsilon}{2}>0$.
    Thus we have:

    $\displaystyle \mid S_{n}-L\mid<\frac{\epsilon}{2}$
    $\displaystyle 2\cdot\mid S_{n}-L\mid<2\cdot\frac{\epsilon}{2}$
    $\displaystyle \mid 2S_{n}-2L\mid<\epsilon$

    By definition $\displaystyle 2S_{n}$ converges to $\displaystyle 2L$ and therefore the limit $\displaystyle \lim_{n\to \infty}2S_{n}$ exists.
    Last edited by detalosi; Aug 3rd 2019 at 03:40 AM.
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