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Math Help - [SOLVED] Infinite series convergence

  1. #1
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    [SOLVED] Infinite series convergence

    Does \sum^\infty _{n=1} \sqrt[n]{2}-1 converge?
    I am sure that \sum^\infty _{n=1} \sqrt[n]{2} diverges, since
    \lim \limits_{n \to \infty} \sqrt[n]{2}=1, but \lim \limits_{n \to \infty} \sqrt[n]{2}-1=0....
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  2. #2
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    \sqrt[n]{2}-1\to0 and \frac{\ln2}n\to0 as n\to\infty, hence \frac{\ln 2}{n}>\frac{1}{n} and your series diverges.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \sqrt[n]{2}-1\to0 and \frac{\ln2}n\to0 as n\to\infty, hence \frac{\ln 2}{n}>\frac{1}{n} and your series diverges.

    I am confused, how does \frac{\ln 2}{n}>\frac{1}{n} imply that \sum^\infty _{n=1} (\sqrt[n]{2}-1)diverges? by comparison?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akolman View Post
    Does \sum^\infty _{n=1} \sqrt[n]{2}-1 converge?
    I am sure that \sum^\infty _{n=1} \sqrt[n]{2} diverges, since
    \lim \limits_{n \to \infty} \sqrt[n]{2}=1, but \lim \limits_{n \to \infty} \sqrt[n]{2}-1=0....
    Note that

    \begin{aligned}\forall{n}\in\mathbb{R}~\sqrt[n]{2}&=2^{\frac{1}{n}}\\<br />
&=e^{\frac{\ln(2)}{n}}\\<br />
&=1+\frac{\ln(2)}{n}+\cdots\end{aligned}

    So

    \begin{aligned}\forall{n}\in\mathbb{R}~\sqrt[n]{2}-1&=1+\frac{\ln(2)}{n}+\cdots-1\\<br />
&=\frac{\ln(2)}{n}+\cdots\geqslant\frac{\ln(2)}{n}  \end{aligned}


    So

    \sum_{n=1}^{\infty}\frac{1}{n}\to\infty\leqslant\s  um_{n=1}^{\infty}\frac{\ln(2)}{n}

    So your series is greater than a divergent series, thus divergent.
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