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Thread: How to prove that a limit exists?

  1. #1
    Member Pranas's Avatar
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    How to prove that a limit exists?

    Hello.

    I have to prove that an actual limit exists while n is approaching infinity.
    I do NOT need to get to the exact limit itself.

    1. It is easy to understand (and prove), that sequence never decreases. That's already half of the job.
    2. I need to make some manipulations that would lead to any specific number being greater than my sequence (irrespective of natural n value).



    I get no further than this, a remaining product is a real headache.
    Thank you for ideas and sorry for possibly broken English
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  2. #2
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    Quote Originally Posted by Pranas View Post
    Hello.

    I have to prove that an actual limit exists while n is approaching infinity.
    I do NOT need to get to the exact limit itself.

    1. It is easy to understand (and prove), that sequence never decreases. That's already half of the job.
    2. I need to make some manipulations that would lead to any specific number being greater than my sequence (irrespective of natural n value).



    I get no further than this, a remaining product is a real headache.
    Thank you for ideas and sorry for possibly broken English
    $\displaystyle \lim\limits_{n\to\infty}\prod\limits^n_{k=1}\frac{ 2^k+1}{2k}=\prod\limits^\infty_{k=1}\left(1+\frac{ 1}{2^k}\right)$ . Applying logarithm you get that the infinite product converges

    iff the infinite series $\displaystyle \sum\limits^\infty_{k=1}\log\left(1+\frac{1}{2^k}\ right)$ converges (the logarithm's base never minds, of course).

    Lemma: if $\displaystyle \{a_n\}$ is a positive sequence, the series $\displaystyle \sum\limits^\infty_{k=1}\log\left(1+a_k\right)$ converges iff the series $\displaystyle \sum\limits^\infty_{k=1}a_k$ converges.

    Proof sketch: assuming $\displaystyle a_k\xrightarrow [k\to\infty]{}0$ , we get what we want from the limit test for positive series, since $\displaystyle \frac{\log(1+a_k)}{a_k}\xrightarrow [k\to\infty]{}1\,\,\,\,\square$

    Tonio
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  3. #3
    Member Pranas's Avatar
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    Thank you very much, I was not aware of the lemma.
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