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Math Help - Prove, if a sequence converges, its subsequences converge and have the same limit?

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    Prove, if a sequence converges, its subsequences converge and have the same limit?

    Prove that if <a> converges, then every subsequence of <a> converges and has the same limit as a.
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    Quote Originally Posted by qtpipi View Post
    Prove that if <a> converges, then every subsequence of <a> converges and has the same limit as a.


    let \epsilon > 0

    suppose that \{a_n\} \to a

    let \{a_{n_k}\} be any subsequnce

    Since the sequence convereges there is an N \in \mathbb{N} such that for all n > N

    |a_n-a|< \epsilon

    but since \{a_{n_k}\} if futher in the sequnce then

     <br />
\{a_{n}\}<br />

    so |\{a_{n_k}\}-a|< \epsilon
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    Quote Originally Posted by TheEmptySet View Post

    but since \{a_{n_k}\} if futher in the sequnce then

     <br />
\{a_{n}\}<br />

    so |\{a_{n_k}\}-a|< \epsilon
    I'm not sure what you're saying here, can you rephrase it? Thanks
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    Quote Originally Posted by qtpipi View Post
    I'm not sure what you're saying here, can you rephrase it? Thanks

    The integer n_k \ge n a subsequence has all of the same terms as the regular sequence. It just skips some

    ie a_n=\frac{1}{n} and

    a_{n_k}=\frac{1}{2^{n_k}}

    the first gives the sequence

    a_n=\left\{ \frac{1}{1} ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,... \right\}

    but


    a_{n_k}=\left\{ \frac{1}{1} , \frac{1}{2} ,\frac{1}{4} ,\frac{1}{8} ,\frac{1}{16} ,... \right\}

    the subsequce skips terms from the original so to get the term

    \frac{1}{16} n=16 but n_k=4

    So  n \ge n_k
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