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Thread: [SOLVED] Infinite series convergence

  1. #1
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    [SOLVED] Infinite series convergence

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

    I am confused, how does $\displaystyle \frac{\ln 2}{n}>\frac{1}{n}$ imply that $\displaystyle \sum^\infty _{n=1} (\sqrt[n]{2}-1)$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 $\displaystyle \sum^\infty _{n=1} \sqrt[n]{2}-1$ converge?
    I am sure that $\displaystyle \sum^\infty _{n=1} \sqrt[n]{2}$ diverges, since
    $\displaystyle \lim \limits_{n \to \infty} \sqrt[n]{2}=1$, but $\displaystyle \lim \limits_{n \to \infty} \sqrt[n]{2}-1=0$....
    Note that

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

    So

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


    So

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

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