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!
 A: 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.
A: Hint:
You may use Cauchy condensation test:
$$\sum_{n=2}^{\infty}\frac{1}{n\log n} \sim \sum_{n=2}^{\infty}\frac{2^n}{2^n\log 2^n}=\sum_{n=2}^{\infty}\frac{1}{n\log 2}$$
A: 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.
A: 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 $\alpha>1$.
