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I'm trying to compute ( or at least prove convergence/divergence) of $$\sum_{n=1}^{\infty} (-1)^{n} \frac{n}{\sqrt{n^3+2}}$$

But I can't get the root or ratio test to work..... Nor am I sure how to use the integral test in this scenario. Any hints?

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    The series converges by the alternating series test. – 2017-02-06
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    I feel like this is divergent because each term is approximately equal to $n^{-1/2}$. And then applying the p- test – 2017-02-06
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    $n^{-1/2} \to 0$ and we have an alternating series, this is all that is necessary @Squirtle – 2017-02-06
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    But you have an alternating series. It is therefore enough that the term without $(-1)^n$ goes to $0$ monotonically. See here https://en.m.wikipedia.org/wiki/Alternating_series_test – 2017-02-06
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    Right. . . . i was taking the absolute value first..... Thanks! – 2017-02-06
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    How might we compute the value? Is it even possible? – 2017-02-06

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Conditional convergence: The sequence $a_n:=\frac{n}{\sqrt{n^3+2}}$ is positive, monotonically decreasing and have limit $0$ when $n\to\infty$, so the series $$ \sum_{n=0}^\infty(-1)^n\frac{n}{\sqrt{n^3+2}} $$ is convergent by Leibniz criterion.

Absolute convergence: We need to check if $\sum_{n=0}^\infty a_n$ is convergent. It is not. To prove it notice that $a_n\sim n^{-1/2}$ and conclude with the limit comparison test since the series $\sum_nn^{-1/2}$ is divergent (by the $p$-series test).

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    I'm not sure why it's monotonically decreasing. – 2017-02-06
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    Okay..... Never mind. I see why. Take derivative and compute the root. – 2017-02-06