Let $f(x)=\dfrac{1}{x\ln x}\quad(x>1)$
- Calculate the General Euler constant $$L=\lim_{n\to\infty}\left(\sum_{k=2}^nf(k)-\int_2^nf(x)dx\right)$$
- Estimate the order of $\{x_n\}$ which is the form $$x_n=\sum_{k=2}^nf(k)-\int_2^nf(x)dx-L$$
- Find the constant $c$ such that the order of $$y_n=\sum_{k=2}^n(k)-\int_2^{c+n}f(x)dx-L$$ is greater than that of $\{x_n\}$,and find the order of $\{y_n\}$
What I've tried is trivial.
By the convex property and Integral mean value theorem,
$$
\int_k^{k+1}f(x)dx\in \left(f(k+1),\frac{1}{2}\big(f(k)+f(k+1)\big)\right),
$$ sum by $k$ from $2$ to $n$, hence we have $$\int_2^{n+1}f(x)dx-f(n+1)+f(2)>\sum_{k=2}^nf(k)>\int_2^{n+1}f(x)dx-\frac{1}{2}f(n+1)+\frac{1}{2}f(2)$$
which implies
$$\sum_{k=2}^nf(k)-\int_2^nf(x)dx\in\left(\int_n^{n+1}f(x)dx-\frac{1}{2}f(n+1)+\frac{1}{2}f(2),\int_n^{n+1}f(x)dx-f(n+1)+f(2)\right)$$
What is obvious is that $\int_a^bf(x)dx=\ln(\ln b)-\ln(\ln a)$.The existence of $L$ is easy and then from the inequality above, the conclusion is $L\in[\frac{1}{2}f(2),f(2)]$ ,which is far away from what the problem want us to do.I wonder if it is possible to get an explicit answer of $L$ and if so, what form should it be？
