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I could not find how I can show that the following series is convergent or not.


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You could use, with a bit of work, the Alternating Series Test. – David Mitra Dec 25 '12 at 17:01
Note that the series is not absolutely converges. – Babak S. Dec 25 '12 at 17:20

Hint: Use Dirichlet test for series convergence.

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Quote: A particular case of Dirichlet's test is the more commonly used alternating series test for the case... – Did Dec 25 '12 at 18:44

$\log(n)=3\log(n^{\tfrac{1}{3}}) \leq 3(n^{\tfrac{1}{3}}-1)$ so the terms of this alternating series converge to zero and

$$\frac{\log(n+1)}{\sqrt{n+1}}\leq\frac{\log(n)+\frac{1}{n}}{\sqrt{n}+\frac{1}{2\sqrt{n+1}}} = \frac{\log(n)}{\sqrt{n}} - \frac{\log(n)-2\sqrt{1+1/n}}{\textrm{something positive}}$$

which shows that the terms are eventually decreasing. Therefore the series converges.

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Of course, one also needs that the sequence $(\ln (n)/\sqrt n)$ is eventually decreasing (which it is by, e.g., a derivative analysis). – David Mitra Dec 25 '12 at 17:18
@DavidMitra The inequality for $\log$ ensures that the terms converge to zero no? Nothing more is needed. – WimC Dec 25 '12 at 17:25
Maybe I'm missing something; but here you need to show that $(\ln n/\sqrt n)$ is decreasing and has limit zero. The terms converge to $0$ by your inequality; but I don't see how it implies that $(\ln n/\sqrt n)$ is decreasing (and decreasing is needed in the Alternating Series Test: consider $\sum_{n=1}^\infty (-1)^n a_n$ where $a_n=1/n$ for $n$ odd and $a_n=1/n^2$ for $n$ even). – David Mitra Dec 25 '12 at 17:31
@DavidMitra Excellent point. :-) – WimC Dec 25 '12 at 17:39

WimC has shown that indeed $\frac{\log n}{\sqrt{n}}$ goes to $0$ as $n\to \infty$. But it's also necessary to show that the sequence $\frac{\log n}{\sqrt{n}}$ is decreasing, or at least decreasing after a certain value of $n$. To do this, you can take the derivative of $\frac{\log n}{\sqrt{n}}$ with respect to $x$, which is: $\frac{2-\log x}{2x^{3/2}}$ after some simplification. The denominator is clearly positive, but after $\log x > 2$ the numerator is less than 0, so the derivative is clearly negative. This means that when $n > e^2$, the sequence will be decreasing. Hence the series, by the alternating series test, is convergent.

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