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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the infinite series of $(-1)^n/(nln(n))$ for $n = 2,3,4,\ldots$ to infinity, is the series conditionally convergent, absoultely convergent, or divergent?

I took two approaches to solve this problem; First, using the absolute value test, the series can now undergo the integral test, which comes out at as $$\ln(\ln(\infty))-\ln(\ln(2)=\infty,$$ therefore the series is not absolutely convergent. The alternating series test can then be applied to the series which, since the series is both approaching zero as $n$ approaches infinity and the and the series without the alternating signs is non-increasing for all $n$, shows that it is conditionally convergent. This is the answer in the back of the textbook.

The second method that I decided to take was to do a direct comparison; after the absolute value of the series has been taken (testing for absolute convergence), I compared the series $1/(n\ln(n))$ with the p-series $1/(n^{1.1})$. The p-series is greater than $1/(n\ln(n))$ for all $n\ge 4$. So I expanded the p-series out to show that the first 3 terms in the sequence plus the series of $(1/(n^{1.1}))$ from 4 to infinity equaled the p-series summed from 1 to infinity. This showed that the series $(1/(n^{1.1}))$ from 4 to infinity is convergent.

Now using the Direct Comparison Test, you can show that the series of $(1/(n^{1.1}))$ from 4 to infinity is greater than or equal to the series $1/(n\ln(n))$ from 4 to infinity. Thusly, the latter series must converge since it is smaller than the p-series.

The series stated in the problem can then be written as the 'n=2' term plus the 'n=3' plus the series of $1/(n\ln(n))$ from 4 to infinity. Since it was just proven that the series of $1/(n\ln(n))$ from 4 to infinity is convergent through the direct comparison test with a larger, and convergent series, it can be said that something convergent plus two constants (the n=2 term and n=3 term of $1/(n\ln(n)))$ will still be finite and thusly convergent. Thus, through the direct comparison test, the series of $1/(n\ln(n))$ with an index of 2 to infinity is convergent and the original alternating series must, thusly, be absoultely convergent.

My teacher and my entire Calc II class spent the entire period going over this issue today. The integral test of the series seems to disagree with the comparison test. Does anyone see some issue with the line of logic here or a technicality about the tests used to prove the absolute/conditional convergence of the series? Or does anyone even know what the correct answer would be for this problem?

share|cite|improve this question

It is incorrect that $$\sum_{n=4}^{k} \dfrac1{n \log n} < \sum_{n=4}^{k} \dfrac1{n^{1.1}}$$ for all $k$. In fact only the reverse is true, i.e., for any $\epsilon>0$, there exists $k$ such that $$\sum_{n=4}^{k} \dfrac1{n \log n} > \sum_{n=4}^{k} \dfrac1{n^{1+ \epsilon}}$$

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.

You may want to look here for a related question. Convergence of the series $\sum \limits_{n=1}^{\infty} \frac{1}{n\log^s n}$

share|cite|improve this answer

Your Answer


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.