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We were having a discussion in class yesterday about why the infinite sum of $$\sum_{n=2}^{∞}\frac{1}{n\ln(n)}$$ is divergent.

My friend tried to show that it converges by using the direct comparison test with the convergent p-series $n^{-1.00000001}$ which is greater than $\frac{1}{n\ln(n)}$. for a very long time, but not everywhere. I was trying to come up with a rigorous proof for why that series was actually less than $\frac{1}{n\ln(n)}$ and I got this far. What I was wondering is if the use of limits is valid in this context:

$$\frac{1}{n\ln(n)} > \frac{1}{n^{1+(1/a)}}\text{where }a>1$$

$$n\ln(n) < n^{1+(1/a)} \longrightarrow \frac{n\ln(n)}{n^{1+(1/a)}} < 1$$

$$\lim_{n\to\infty} \frac{n\ln(n)}{n^{1+(1/a)}}<1 \longrightarrow \lim_{n\to\infty} \frac{\frac{1}{n}}{\frac{1}{a}(\frac{1}{a}+1)n^{\frac{1}{a}-1}}<1 $$ (L’Hospital’s rule twice)

$$\lim_{n\to\infty} \frac{a}{n(\frac{1}{a}+1)n^{(\frac{1}{a}-1)}}<1 \longrightarrow \lim_{n\to\infty} \frac{a}{(\frac{1}{a}+1)n^{\frac{1}{a}}} = 0 < 1$$

therefore $$\sum_{n=1}^{\infty}\frac{1}{n\ln(n)} > \sum_{n=1}^{\infty}\frac{1}{n^{1+\frac{1}{a}}}$$ so we cannot conclude that converges because the p-series converges.

Thanks in advance!

Math Attack
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Jim
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    Are we allowed to use the integral test? 1 / (x log x) is a decreasing function, and its antiderivative log(log x) diverges albeit quite slowly. – Noam D. Elkies Mar 25 '25 at 14:58
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    Cauchy condensation test is good for series of this form. – Wojowu Mar 25 '25 at 14:59
  • Yes the integral test was used by someone else in my class, but I was wondering how to refute the attempt to use the direct comparison test. Is the use of limits justified to do this in this case? Thanks! – Jim Mar 25 '25 at 15:06
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    see https://math.stackexchange.com/q/2759494/87355 – Carlo Beenakker Mar 25 '25 at 15:15
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    The series as written diverges trivially, check $n=1$ :) – Severin Schraven Mar 25 '25 at 15:18
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    of course, it should start from $n=2$, I corrected it. – Carlo Beenakker Mar 25 '25 at 15:51
  • $$\sum 2^na_{2^n}=\frac{1}{\ln 2}\sum \frac{1}{n}$$ – Bowei Tang Mar 25 '25 at 16:57
  • Thanks for all the answers. I understand that the series must diverge; I was wondering if the work I did (minus the incorrect initial values for n) is sufficient to justify that $\sum_{n=2}^{\infty}\frac{1}{n\ln(n)} > \sum_{n=2}^{\infty}\frac{1}{n^{1+\frac{1}{a}}}$ Having to use limits made me unsure. – Jim Mar 26 '25 at 00:22

3 Answers3

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If we took the sum at $n=0$, the sum doesn't exist due to $2$ singularities at $n=1$ and $n=0$ (meaning the limit at those points doesn't exist either). To check if it is divergent for $n>1$, let's use the integral test. $$\sum_{n=2}^{\infty} \frac{1}{n \ln (n)}$$ From this, we get.

$$\int_{2}^{\infty} \frac{dx} {x\ln(x)}$$

To see if the series diverges or converges. Do $u$-substitution. If you compute it correctly, the series should diverge since its integral also diverges.

Note: comparison test will not work on this summation.

Bowei Tang
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DAVID CHEDESTER
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Since we know that $$\int \frac{dn}{n \log (n)}=\log (\log (n))$$ we can use with the simplest form of Euler-Maclaurin summation formula. It would give $$S_m=\sum_{n=2}^{m} \frac{1}{n \log (n)}=\log (\log (m))+C+\frac{1}{2 m \log (m)}+O\left(\frac{1}{m^2 \log (m)}\right)$$ For this level of expansion $C\sim \frac 45$.

Using this very truncated expression, for $m=10^6$, it would give $3.41806$ while the summation would give $3.42047$ (relative error $=0.07$%).

Claude Leibovici
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Here is a non calculus solution to show it grows arbitarily large.

for any natural number $N$ .

$\sum_{n=2}^{N} \frac{1}{n\log(n)} \sim \sum_{n=2}^{N} \frac{1}{n \lfloor \log(n) \rfloor}$

we may assume we have log base 2[all logs are proportional so it dosent matter] .

now set $k=\lfloor \log (n) \rfloor$ and analyze the case $N=2^{t}-1$ for some large integer $t$

so $\sum_{n=2}^{2^t-1} \frac{1}{n \lfloor \log(n) \rfloor}=\sum_{k=1}^{t-1} \frac{1}{k} \sum_{i=2^k}^{2^{k+1}-1} \frac{1}{i} \geq \sum_{k=1}^{t-1} \frac{1}{k} \frac{2^{k}}{2^{k+1}-1} \geq \frac{1}{2} \sum_{k=1}^{t-1} \frac{1}{k}$.

And we know harmonic series diverges to infinity.

Lucid
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