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?
Answer
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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