# Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$

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?

• Compare it to the integral $\int_2^\infty \frac{dt}{t\log t}$. – Daniel Fischer Nov 20 '13 at 11:44
• Please be careful with the parenthesis : is it $\frac{1}{n \log n}$ or $\frac{\log n}{n}$? – user37238 Nov 20 '13 at 11:48
• @user37238,Thanks! It is the first for me to use the Mathematics!I am a undergraduate student from China. – abc Nov 20 '13 at 11:53
• I edited the answer following the second formula. Waiting for the OP to better specify his intentions, though. – Avitus Nov 20 '13 at 11:53
• – Martin Sleziak Oct 31 '15 at 20:30

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.

• ,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! – abc Nov 21 '13 at 14:01
• 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. – Ido Sep 6 '16 at 6:09

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.

• I think this may have just saved me for my upcoming exam! Thanks ;) – RGS Jan 24 '17 at 10:59

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.

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

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

and

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

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.