Makenna Booker

2022-07-15

Convergence of series minus logarithm
im trying to solve this problem since two, three days.. Is there someone who can help me to solve this problem step by step. I really want to understand & solve this!

The sum looks like the harmonic series.
My thoughts were to compare this sum with an integral, the lower and the upper riemann-sum, to get an inequation.

iljovskint

Expert

Comparing sum and integral, we get
$\begin{array}{rl}\frac{1}{n\mathrm{log}\left(n\right)}+{\int }_{2}^{n}\frac{\mathrm{d}x}{x\mathrm{log}\left(x\right)}& \le \sum _{k=2}^{n}\frac{1}{k\mathrm{log}\left(k\right)}\\ \text{(1)}& & \le \frac{1}{2\mathrm{log}\left(2\right)}+\frac{1}{3\mathrm{log}\left(3\right)}+{\int }_{3}^{n}\frac{\mathrm{d}x}{x\mathrm{log}\left(x\right)}\end{array}$
Thus,
$\begin{array}{rl}\frac{1}{n\mathrm{log}\left(n\right)}-\mathrm{log}\left(\mathrm{log}\left(2\right)\right)& \le \sum _{k=2}^{n}\frac{1}{k\mathrm{log}\left(k\right)}-\mathrm{log}\left(\mathrm{log}\left(n\right)\right)\\ \text{(2)}& & \le \frac{1}{2\mathrm{log}\left(2\right)}+\frac{1}{3\mathrm{log}\left(3\right)}-\mathrm{log}\left(\mathrm{log}\left(3\right)\right)\end{array}$
By the Mean Value Theorem and since $\frac{1}{n\mathrm{log}\left(n\right)}$ is decreasing, for some $\kappa \in \left(k-1,k\right)$
$\begin{array}{}\text{(3)}& \frac{1}{k\mathrm{log}\left(k\right)}\le \frac{1}{\kappa \mathrm{log}\left(\kappa \right)}=\mathrm{log}\left(\mathrm{log}\left(k\right)\right)-\mathrm{log}\left(\mathrm{log}\left(k-1\right)\right)\end{array}$
Therefore, the red difference in $\left(2\right)$ is decreasing in $n$ since
$\begin{array}{rl}\phantom{\rule{-1cm}{0ex}}& \sum _{k=2}^{n}\frac{1}{k\mathrm{log}\left(k\right)}-\mathrm{log}\left(\mathrm{log}\left(n\right)\right)\\ \text{(4)}& & =\frac{1}{2\mathrm{log}\left(2\right)}-\mathrm{log}\left(\mathrm{log}\left(2\right)\right)+\sum _{k=3}^{n}\left(\frac{1}{k\mathrm{log}\left(k\right)}-\left[\mathrm{log}\left(\mathrm{log}\left(k\right)\right)-\mathrm{log}\left(\mathrm{log}\left(k-1\right)\right)\right]\right)\end{array}$
and by $\left(3\right)$, each blue term in $\left(4\right)$ is negative.
So, $\left(2\right)$ says that the red difference is decreasing and bounded below by $-\mathrm{log}\left(\mathrm{log}\left(2\right)\right)$. Thus, the limit of the red difference exists and is at least $-\mathrm{log}\left(\mathrm{log}\left(2\right)\right)\doteq 0.366512920581664$
Furthermore, $\left(2\right)$ also says that for each $n\ge 3$, the red difference is at most $\frac{1}{2\mathrm{log}\left(2\right)}+\frac{1}{3\mathrm{log}\left(3\right)}-\mathrm{log}\left(\mathrm{log}\left(3\right)\right)\doteq 0.930712768370062$

Jaxon Hamilton

Expert

The function $f:x↦\frac{1}{x\mathrm{log}x}$ is continuous non negative decreasing on $\left[2,+\mathrm{\infty }\right)$ hence the sequence
${S}_{n}=\sum _{k=3}^{n}\left(\frac{1}{k\mathrm{log}k}-{\int }_{k-1}^{k}\frac{dx}{x\mathrm{log}x}\right)$
is convergent, in fact
$f\left(n\right)-f\left(2\right)=\sum _{k=3}^{n}\left(f\left(k\right)-f\left(k-1\right)\right)\le {S}_{n}\le 0$
Can you take it from here?

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