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I need to determine whether the following sequence converge or not.

$$ \sum_{k=2}^{\infty}(-1)\ ^ n \dfrac{ \sqrt[n]{n}}{\ln(n)} $$

I tried to show that this is a leibniz series , but couldn't manage to show that $a_n$ is monotonic decreasing.

Thanks for helping.

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Based on Taylor expansion of the main term, to decompose the main series into several series to analyze separately (with Leibniz for the first, and comparison for the last).

One can write, for $n\to\infty$, $$ \frac{\sqrt[n]{n}}{\ln n} =\frac{ e^{\frac{\ln n}{n}}}{\ln n} =\frac{1+\frac{\ln n}{n}+\frac{\ln^2 n}{2n^2}+o\left(\frac{\ln^2 n}{n^2}\right)}{\ln n} $$ so that $$ (-1)^n\frac{\sqrt[n]{n}}{\ln n} = \frac{(-1)^n}{\ln n}+\frac{(-1)^n}{n}+\frac{(-1)^n\ln n}{2n^2}+o\left(\frac{\ln n}{n^2}\right) $$ This allows you to conclude by the alternating criterion: $\sum_n \frac{(-1)^n}{\ln n}$ and $\sum_n \frac{(-1)^n}{n}$ are conditionally convergent; and the remaining term, $\sum_n \left(\frac{(-1)^n\ln n}{2n^2}+o\left(\frac{\ln n}{n^2}\right)\right)$, is absolutely convergent (e.g., by comparison).

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  • $\begingroup$ The rationale here is: "when Leibniz/alternating series criterion does not apply directly because the term of the series is not monotonic or does not seem so, doing a Taylor series expansion often help to decompose into two sorts of series: the ones on which Leibniz does apply straightforwardly, and the last one which is nicely absolutely convergent." $\endgroup$ – Clement C. May 12 '16 at 18:44
  • $\begingroup$ Thanks for the detailed explanation This was really helpful ! $\endgroup$ – user335501 May 12 '16 at 19:01
  • $\begingroup$ One question - In the Taylor series you made for $e \ ^ {ln(n)/n} $ How could you made this series around 0 ? your function doesn't defined there @Clement C. $\endgroup$ – user335501 May 12 '16 at 19:06
  • $\begingroup$ @Liad Take $x_n=\frac{\ln n}{n}$. Then $x_n \xrightarrow[n\to\infty]{}0$, so you only need to do a Taylor expansion of $e^x$ around $0$: $e^{x_n} = 1+x_n + \frac{x_n^2}{2} + o(x_n^2)$. (The expansion is in $x_n$ at $0$, not in $n$) $\endgroup$ – Clement C. May 12 '16 at 19:08
  • $\begingroup$ Got it, again, thanks ! $\endgroup$ – user335501 May 12 '16 at 19:10
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Let $f(x)=\frac{\ln x}{x}$ then $f'(x)=\frac{1-\ln x}{x^2}\le0$ for all $x\ge e$ hence the sequence $\sqrt[n]n$ is decreasing for all $n\ge3$ and hence $\frac{\sqrt[n]n}{\ln n} $ is also decreasing for $n\ge3$. Now use the Leibniz rule to conclude the convergence.

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  • $\begingroup$ (Leibniz rule = alternated series test) $\endgroup$ – reuns May 12 '16 at 18:08
  • $\begingroup$ @user296113 Can you explain to me why from f(x) you conclude that $\sqrt[n]{n}$ is decrreasing ? $\endgroup$ – user335501 May 12 '16 at 18:11
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    $\begingroup$ @Liad $\sqrt[n]{n}=e^{\ln n/n}$ $\endgroup$ – Wojowu May 12 '16 at 18:12
  • $\begingroup$ @Wojowu okay, got it, thanks ! $\endgroup$ – user335501 May 12 '16 at 18:13
  • $\begingroup$ This shows $f(x)$ is decreasing, but why is $\frac{\exp(f(x))}{\ln(x)} $ also decreasing? $\endgroup$ – darksky May 12 '16 at 18:20
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A quick way to verify your hypothesis is to plot it out. Below you can see, it is monotonic and decreasing (seems to zero). You can then put your effort in this direction.

enter image description here

  1. the function is monotonically decreasing $f(x)=\frac{x^{\frac1x}}{\ln x}$ or $f'(x)<0$
  2. it's limit is zero, i.e. $\lim_{x \to \infty} f(x) =0$

For the first point, let $g(x)=\ln(f(x))=\frac1x\ln x-\ln(\ln x)$, then $$g'(x)=\frac{f'(x)}{f(x)}=\frac {\ln(x)-x-(\ln(x))^2}{x^2\ln x}$$ We know $$\ln(x)\lt x$$ for large x.So $f'(x) \lt 0 $ and thus is monotonically decreasing.

For the second point, because $$\lim_{x \to \infty}\frac1x\ln x=0$$ then $$\lim_{x \to \infty}g(x)=\lim_{x \to \infty}(\frac1x\ln x-\ln(\ln x))=-\infty$$ thus, $$\lim_{x \to \infty}\ln(f(x))=-\infty$$ and $$\lim_{x \to \infty}f(x)=0$$

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