Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$\sum_{n=2}^\infty(-1)^n\frac{\ln(n)}{\sqrt{n}}$$

share|improve this question
1  
You could use, with a bit of work, the Alternating Series Test. –  David Mitra Dec 25 '12 at 17:01
1  
Note that the series is not absolutely converges. –  Babak S. Dec 25 '12 at 17:20

3 Answers 3

Hint: Use Dirichlet test for series convergence.

share|improve this answer
    
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.

share|improve this answer
    
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
3  
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.