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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?

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6 Answers 6

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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.

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

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  • $\begingroup$ ,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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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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.

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

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

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

Now, for $x>1$,

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

$=\frac{1}{2}(\frac{1}{2^{x-1}}+\frac{1}{4^{x-1}}+....)$

$=\frac{1}{2^x}(\frac{1}{1-\frac{1}{2^{x-1}}})$

$=\frac{1}{2}(\frac{1}{2^{x-1}-1})$.

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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