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I would like to see whether or not $$\sum\limits_{n=1}^{\infty}(-1)^{n}\dfrac{\ln(n)}{\sqrt{n}}$$ is a convergent series.

Root test and ratio test are both inconclusive. I tried the alternating series test after altering the form of the series: $$\sum\limits_{n=1}^{\infty}(-1)^{n-1}\left[\dfrac{-\ln(n)}{\sqrt{n}}\right]\text{.}$$ After using L-Hospital, it's clear that $\lim\limits_{n \to \infty}\left[\dfrac{-\ln(n)}{\sqrt{n}}\right] = 0$. To show that it's decreasing led to me finding the derivative $\dfrac{\ln(n)}{2n^{3/2}}-\dfrac{1}{n^{3/2}}$, which I could set to be less than $0$, but a plot has shown that $n < e^{2}$ is not where $\dfrac{-\ln(n)}{\sqrt{n}}$ is decreasing.

So all that remains is a comparison test. I can't think of a clever comparison to use for this case. Any ideas?

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  • $\begingroup$ Use Dirichlet test. I think alternating test is also works. $\endgroup$ – Hanul Jeon Oct 21 '14 at 5:08
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    $\begingroup$ Continue with your alternating series test work. You can show that after a while the function $\frac{\log t}{t^{1/2}}$ is decreasing, and that is good enough. $\endgroup$ – André Nicolas Oct 21 '14 at 5:09
  • $\begingroup$ This is $-\eta'\bigg(\dfrac12\bigg).~$ See Dirichlet $\eta$ function for more information. $\endgroup$ – Lucian Oct 21 '14 at 6:03
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If $f(x)=\dfrac{\ln x}{\sqrt{x}}$ then

$$f'(x) = \dfrac{2-\ln x}{2\sqrt{x^3}}.$$

Which is negative for all $x>e^2$. So $f(n)=b_n$ is decreasing for all integers $n\ge 8$.

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$\dfrac{\ln(n)}{\sqrt{n}}$ is mono-tonic decreasing after $n=\lceil e^2 \rceil$ and remains bounded between $n=1$ to $\lfloor e^2 \rfloor$, so from alternating series test $\sum\limits_{n=1}^{\infty}(-1)^{n}\dfrac{\ln(n)}{\sqrt{n}}$, must converge.

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You can use Leibniz test:

if you have a series like $\sum\limits_{n=1}^{\infty}(-1)^{n}{a_n}$ and ${a_n}$ is decrescent and infinitesimal when $ n \rightarrow \infty$, then the series

$\sum\limits_{n=1}^{\infty}(-1)^{n}{a_n}$ converges

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