# Calculus 1: limit of sum

I'm studying for my calculus 1 exam and came across this sample question from the professor's collection:

Calculate: $$\lim\limits_{n\ \rightarrow\ \infty} \frac{1}{2\log(2)}+\frac{1}{3\log(3)} + \dots + \frac{1}{n\log n}$$ (hint: separate into blocks)

Unfortunately the sample questions don't include answers and I'm at a loss as to how to proceed; I'd really appreciate some help.

Thanks!

• Separate $2^j \le n< 2^{j+1}$ for $j=1,2,\ldots$. – Song Jan 29 '19 at 12:57
• @Song meaning I should separate into blocks: $\frac{1}{2^j\log(2^j)}+\dots+\frac{1}{(2^{j+1}-1)\log(2^{j+1}-1)}$ ? – Or Bairey-Sehayek Jan 29 '19 at 13:23
• Yes, it can be a good choice. In fact, it is in the same spirit of Cauchy condensation test. One can show $\frac{1}{2^j\log (2^j)}+\cdots +\frac{1}{(2^{j+1}-1)\log(2^{j+1}-1)}\ge \frac{c}{j}$ for some $c>0$. – Song Jan 29 '19 at 13:27

Hint:

You may use Cauchy condensation test: $$\sum_{n=2}^{\infty}\frac{1}{n\log n} \sim \sum_{n=2}^{\infty}\frac{2^n}{2^n\log 2^n}=\sum_{n=2}^{\infty}\frac{1}{n\log 2}$$

Hint. Note that $$\frac{1}{2\log(2)}+\frac{1}{3\log(3)} + \dots + \frac{1}{n\log n}\geq \int_2^{n+1}\frac{dx}{x\log(x)}$$ where the left-hand side is the sum of areas of $$n-1$$ rectangles which "dominates" the area given by the integral on the right-hand side.

To flesh out the hint "separate into blocks" somewhat:

\begin{align} {1\over3\log3}+{1\over4\log4} &\gt{1\over4\log4}+{1\over4\log4}\\ &={2\over4\log(2^2)}\\ &={1\over2\log2}\cdot{1\over2}\\ {1\over5\log5}+{1\over6\log6}+{1\over7\log7}+{1\over8\log8} &\gt{1\over8\log8}+{1\over8\log8}+{1\over8\log8}+{1\over8\log8}\\ &={4\over8\log(2^3)}\\ &={1\over2\log2}\cdot{1\over3} \end{align}

etc.

From $$x-1\le\lfloor x\rfloor\le x,$$ we draw

$$\frac1{x\log x}\le\frac1{\lfloor x\rfloor\log\lfloor x\rfloor}\le\frac1{(x-1)\log(x-1)}.$$

Then, integrating from $$3$$ to $$n+1$$,

$$\int_3^{n+1}\frac{dx}{x\log x}\le\sum_{k=3}^n\frac1{k\log k}\le\int_3^{n+1}\frac{dx}{(x-1)\log(x-1)}$$

or

$$\log\log(n+1)-\log\log3\le\sum_{k=3}^n\frac1{k\log k}\le\log\log n-\log\log2.$$

This clearly shows that the sum is asymptotic to $$\log\log n$$.

By the same method,

$$\sum_{k=3}^n\frac1{k\log^\alpha k}\sim\frac1{(1-\alpha)\log^{\alpha-1}n}$$ converges for $$\alpha>1$$.