Recently, I encountered a problem about infinite series. So my question is how to know whether the infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is convergent?


6 Answers 6


Check the conditions fit for the Condensation Test :

$$a_n:=\frac1{n\log n}\implies 2^na_{2^n}=\frac{2^n}{2^n\log2^n}=\frac1{\log 2}\frac1n$$

and since the series of the rightmost sequence is just a multiple of the harmonic series and thus diverges, also our series diverges.

  • 1
    $\begingroup$ I think this may have just saved me for my upcoming exam! Thanks ;) $\endgroup$
    – RGS
    Jan 24, 2017 at 10:59

To see whether $\sum_2^\infty 1/(n \log n)$ converges, we can use the integral test. This series converges if and only if this integral does: $$ \int_2^\infty \frac{1}{x \log x} dx = \left[\log(\log x)\right]_2^\infty $$ and in fact the integral diverges.

This is part of a family of examples worth remembering. Note that $$ d/dx \log(\log(\log x)) = d/dx \log(\log x) \cdot \frac{1}{\log (\log x)} = \frac{1}{x \log x \log(\log x)} $$ and $\log (\log (\log x)) \to \infty$ as $x \to \infty$ hence $\sum \frac{1}{n \log n \log (\log n)}$ diverges as well. Similarly, by induction we can put as many iterated $\log$s in the denominator as we want (i.e. $\sum \frac{1}{n \log n \log(\log n) \ldots \log (\ldots (\log n) \ldots )}$ where the $i$th log is iterated $i$ times), and it will still diverge. However, as you should check, $\sum \frac{1}{x \log^2x}$ converges, and in fact (again by induction) if you square any of the iterated logs in $\sum \frac{1}{n \log n \log(\log n) \ldots \log (\ldots (\log n) \ldots )}$ the sum will converge.

  • $\begingroup$ ,Thanks for your detailed solutioin.I know the series can be see as a area of trapezoid with curved edge that the delta x=1,so the sum[1/(nLn(n)),{n,2,Infinity}] is less than the integrate[1/(xLn(x),{x,2,Infinity}].However,I cannot understand why the latter integral converge can reduct the former series converge?Can you tell me why?Thanks sincerely! $\endgroup$
    – abc
    Nov 21, 2013 at 14:01
  • 1
    $\begingroup$ Excellent answer. Two comments that might look obvious: 1. It is not only squaring, but also raising to the power of any $c>1$. 2. When you square (or raise to the power of $c>1$), you should do that to an outermost $\log$, or else the series will still diverge. For example, $\sum1/\left(n\log n\log\log^{2}n\right)$ diverges. $\endgroup$
    – Ido
    Sep 6, 2016 at 6:09
  • $\begingroup$ Julien, are you sure about your claim that $\frac{1}{n\log^2 n}$ is summable? $\endgroup$ Nov 8, 2019 at 16:30
  • $\begingroup$ Yes, because the condensation test math.stackexchange.com/a/574528/426834 yields $\sum_n \frac1{(\log 2)^2}\frac1{n^2} = \frac1{(\log 2)^2} \sum_n \frac1{n^2}$, which converges. This is also well known. $\endgroup$ May 11 at 9:49

We circumvent using the integral test or its companion, the Cauchy condensation test. Rather, we use creative telescoping to show that the series $\sum_{n=3}^\infty \frac{1}{n\log(n)}$ diverges. To that end, we now proceed.

We will use the well-known inequalities for the logarithm (SEE THIS ANSWER)

$$\frac{x-1}{x} \le \log(x)\le x-1 \tag1$$

Using the right-hand side inequality in $(1)$, we see that

$$\log\left(\frac{n+1}{n}\right)\le \frac1n \tag 2$$


$$\log\left(\frac{\log(n+1)}{\log(n)}\right)\le \frac{\log(n+1)}{\log(n)}-1 \tag3$$

Applying $(2)$ and $(3)$ yields

$$\begin{align} \sum_{n=3}^N \frac{1}{n\log(n)} &\ge \sum_{n=3}^N \frac{\log\left(\frac{n+1}{n}\right)}{\log(n)}\\\\ &=\sum_{n=3}^N \left(\frac{\log(n+1)}{\log(n)} -1\right)\\\\ &\ge \sum_{n=3}^N \log\left(\frac{\log(n+1)}{\log(n)}\right)\\\\ &=\sum_{n=3}^N \left(\log(\log(n+1)) -\log(\log(n)) \right)\\\\ &=\log(\log(N+1))-\log(\log(3)) \end{align}$$

Inasmuch as $\lim_{N\to \infty}\log(\log(N+1))=\infty$, the series of interest diverges by comparison.

And we are done!

TOOLS USED: The right-hand side inequality in $(1)$ and summing a telescoping series.

  • $\begingroup$ Your answer is good, Mark, it's just a few years late... $\endgroup$ Oct 7, 2019 at 16:56
  • $\begingroup$ @Tony Hi Tony. Much appreciated! $\endgroup$
    – Mark Viola
    Oct 7, 2019 at 17:26

Here is another possible answer.

We will derive the asymptotic formula of the partial sum $\sum_{1< n\leqslant x}\frac{1}{n\log n}$ to show that this series diverges.

By Euler's summation formula, we have $$\begin{align*} \sum_{1< n\leqslant x}\frac{1}{n\log n} &=\frac{1}{2\log 2}+\sum_{2< n\leqslant x}\frac{1}{n\log n}\\ &=\frac{1}{2\log 2}+\int_{2}^x\frac{1}{t\log t}dt-\int_{2}^x(t-[t])\frac{\log t+1}{t^2\log^2t}dt-\frac{x-[x]}{x\log x}\\ &=\log\log x+\frac{1}{2\log 2}-\log\log 2-\int_{2}^xI(t)dt-O\left(\frac{1}{x\log x}\right) \end{align*}$$ where $I(t)=(t-[t])\frac{\log t+1}{t^2\log^2t}$.

Since $$\int_{2}^xI(t)dt=\int_2^\infty I(t)dt-\int_x^\infty I(t)ft,$$ and $$I(t)\leqslant \frac{\log t+1}{t^2\log^2t},$$ we obtain ($x\geqslant 2$) $$\int_x^\infty I(t)dt\leqslant\int_x^\infty \frac{\log t+1}{t^2\log^2t}=O\left(\frac{1}{x\log x}\right)$$ and $$\int_2^\infty I(t)dt\leqslant\int_2^\infty \frac{\log t+1}{t^2\log^2t}=\frac{1}{2\log 2}.$$

This implies $-\int_{2}^xI(t)dt=C+O\left(\frac{1}{x\log x}\right)$ (the first integral is not related to $x$ and it's bounded by $1/2\log 2$, thus it's a constant).

Therefore, we have $$\sum_{1< n\leqslant x}\frac{1}{n\log n}=\log\log x+B+O\left(\frac{1}{x\log x}\right)$$

This tells us, as $x\to\infty$, the sum goes to infinity too.

I understand that this asymptotic formula is not needed to prove its divergence, but it may provide you with another way of proving convergence or divergence.


We use integral test to see whether series converge or diverge.

Let $f(x)=\frac{1}{x \log x}.$

Take $u=\log x$ then $du=\frac{1}{x} dx$

Now, $\int f(x) dx=\int \frac{1}{u}du$

$\int_2^\infty f(x)=\log(\log x))|^\infty_2=\lim_{b \rightarrow \infty} \log(\log b)-\log (\log 2)=\infty.$

Hence the series diverges.


$\sum_{n≥2} \frac{1}{n(\log n)}=\sum_{n≥2} \int_{1}^{\infty} \frac{1}{n^x} dx$

$=\int_{1}^{\infty} \zeta_2(x) dx$ where [$\zeta_2(x)=\zeta(x)-1$]

So, this is the area under the curve $\zeta(x), x≥1, y≥0$.

Now, for $x>1$,

$\zeta_2(x) > {\frac{1}{2^x} + 2\frac{1}{4^x}+ 4\frac{1}{8^x} +......}$




So, $\int_{1}^{\infty} \zeta_{2}(x) dx >\int_{1}^{\infty} \frac{1}{2}(\frac{1}{2^{x-1}-1}) dx=\int_{0}^{\infty} \frac{1}{2}(\frac{1}{2^z-1})dz $ which diverges.

This means $\int_{1}^{\infty} \zeta_{2}(x) dx$ also diverges.

And this implies $\sum_{n≥2} \frac{1}{n(\log n)}$ diverges.


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