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When testing to determine the convergence or divergence of series with positive terms, there's a common way by comparing them with other series which we already know converge or diverge.

My question is, how do we choose the proper to-be-compared series? I hope to get some detailed methodology about this. I am a bit confused - do I have to even rely on my intuition?

For instance, how do I choose a comparison series for this given one below:

$$\sum_{n=2}^\infty\frac{1}{n\ln n}$$

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2 Answers 2

Usually, you can choose the highest order of the denominator and numerator, respectively. So in your problem, for the numerator choose $1$, whose highest order is $0$, and for denominator choose $n$. Then, the sequence you should choose is $1/n$.

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But the OP's sequence is smaller than your suggestion, so the divergence of your example tells you nothing about the OP's series... –  Jason DeVito Nov 4 '12 at 4:07
    
it seems that the 1/n can't work out here. –  Aw Qirui Guo Nov 4 '12 at 14:59
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In your case, the convergence of $\displaystyle \sum_{n=2}^{\infty} \dfrac1{n \log n}$ can be checked by using the following convergence test. If we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges.

Note that $$\sum_{n=2^k}^{2^{k+1}-1}\dfrac1{n \log n} > \sum_{n=2^k}^{2^{k+1}-1} \dfrac1{2^k \log \left(2^k \right)} = \dfrac{2^k}{2^k k \log(2)} = \dfrac1{\log 2} \dfrac1k$$

Hence, $$\displaystyle \sum_{n=2}^{\infty} \dfrac1{n \log n} > \dfrac1{\log 2} \sum_{k=1}^{\infty} \dfrac1k$$ Hence, it diverges.

In general, if you want to prove $$\sum_{k=1}^{\infty} a_k$$ diverges and you are unable to find $b_k$ such that $a_k > b_k$ and $$\sum_{k=1}^{\infty} b_k$$ diverges, your next bet is to find $b_k$ such that $\displaystyle \sum_{n=f(k)}^{f(k+1)-1} a_n > b_k$, where $f(k)$ is some strictly monotone increasing function, such that $\sum b_k$ diverges.

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I don't have a means to comment on other people's answers apparently, but isn't $$\displaystyle \sum_{n=2}^{\infty} \dfrac1{n \log n} < \dfrac1{\log 2} \sum_{k=1}^{\infty} \dfrac1k$$ ? –  seth Nov 4 '12 at 5:14
    
@seth: But the right side is unbounded, so this doesn't prove anything. –  marty cohen Nov 4 '12 at 5:33
    
@seth: What I meant was the following. $$\sum_{n=2}^{2^{k+1}-1} \dfrac1{n \log n} > \dfrac1{\log 2} \sum_{n=2}^{k} \dfrac1n$$ The limits of the two are different. –  user17762 Nov 4 '12 at 5:44
    
I know marty, I just had a question about Marvis's answer. @Marvis Thanks! –  seth Nov 4 '12 at 6:44
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