Makenna Booker

Answered

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!

$Show\text{}\mathrm{\exists}\text{}\beta \in [0,1],so\text{}that..\phantom{\rule{0ex}{0ex}}\underset{n\to \mathrm{\infty}}{lim}(\sum _{k=2}^{n}{\displaystyle \frac{1}{k\cdot \mathrm{log}(k)}}-\mathrm{log}(\mathrm{log}(n)))=\beta \text{}\phantom{\rule{0ex}{0ex}}$

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.

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!

$Show\text{}\mathrm{\exists}\text{}\beta \in [0,1],so\text{}that..\phantom{\rule{0ex}{0ex}}\underset{n\to \mathrm{\infty}}{lim}(\sum _{k=2}^{n}{\displaystyle \frac{1}{k\cdot \mathrm{log}(k)}}-\mathrm{log}(\mathrm{log}(n)))=\beta \text{}\phantom{\rule{0ex}{0ex}}$

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.

Answer & Explanation

iljovskint

Expert

2022-07-16Added 18 answers

Comparing sum and integral, we get

$\begin{array}{rl}\frac{1}{n\mathrm{log}(n)}+{\int}_{2}^{n}\frac{\mathrm{d}x}{x\mathrm{log}(x)}& \le \sum _{k=2}^{n}\frac{1}{k\mathrm{log}(k)}\\ \text{(1)}& & \le \frac{1}{2\mathrm{log}(2)}+\frac{1}{3\mathrm{log}(3)}+{\int}_{3}^{n}\frac{\mathrm{d}x}{x\mathrm{log}(x)}\end{array}$

Thus,

$\begin{array}{rl}\frac{1}{n\mathrm{log}(n)}-\mathrm{log}(\mathrm{log}(2))& \le {\sum _{k=2}^{n}\frac{1}{k\mathrm{log}(k)}-\mathrm{log}(\mathrm{log}(n))}\\ \text{(2)}& & \le \frac{1}{2\mathrm{log}(2)}+\frac{1}{3\mathrm{log}(3)}-\mathrm{log}(\mathrm{log}(3))\end{array}$

By the Mean Value Theorem and since $\frac{1}{n\mathrm{log}(n)}$ is decreasing, for some $\kappa \in (k-1,k)$

$\begin{array}{}\text{(3)}& \frac{1}{k\mathrm{log}(k)}\le \frac{1}{\kappa \mathrm{log}(\kappa )}=\mathrm{log}(\mathrm{log}(k))-\mathrm{log}(\mathrm{log}(k-1))\end{array}$

Therefore, the red difference in $(2)$ is decreasing in $n$ since

$\begin{array}{rl}\phantom{\rule{-1cm}{0ex}}& {\sum _{k=2}^{n}\frac{1}{k\mathrm{log}(k)}-\mathrm{log}(\mathrm{log}(n))}\\ \text{(4)}& & =\frac{1}{2\mathrm{log}(2)}-\mathrm{log}(\mathrm{log}(2))+\sum _{k=3}^{n}\left({\frac{1}{k\mathrm{log}(k)}-[\mathrm{log}(\mathrm{log}(k))-\mathrm{log}(\mathrm{log}(k-1))]}\right)\end{array}$

and by $(3)$, each blue term in $(4)$ is negative.

So, $(2)$ says that the red difference is decreasing and bounded below by $-\mathrm{log}(\mathrm{log}(2))$. Thus, the limit of the red difference exists and is at least $-\mathrm{log}(\mathrm{log}(2))\doteq 0.366512920581664$

Furthermore, $(2)$ also says that for each $n\ge 3$, the red difference is at most $\frac{1}{2\mathrm{log}(2)}+\frac{1}{3\mathrm{log}(3)}-\mathrm{log}(\mathrm{log}(3))\doteq 0.930712768370062$

$\begin{array}{rl}\frac{1}{n\mathrm{log}(n)}+{\int}_{2}^{n}\frac{\mathrm{d}x}{x\mathrm{log}(x)}& \le \sum _{k=2}^{n}\frac{1}{k\mathrm{log}(k)}\\ \text{(1)}& & \le \frac{1}{2\mathrm{log}(2)}+\frac{1}{3\mathrm{log}(3)}+{\int}_{3}^{n}\frac{\mathrm{d}x}{x\mathrm{log}(x)}\end{array}$

Thus,

$\begin{array}{rl}\frac{1}{n\mathrm{log}(n)}-\mathrm{log}(\mathrm{log}(2))& \le {\sum _{k=2}^{n}\frac{1}{k\mathrm{log}(k)}-\mathrm{log}(\mathrm{log}(n))}\\ \text{(2)}& & \le \frac{1}{2\mathrm{log}(2)}+\frac{1}{3\mathrm{log}(3)}-\mathrm{log}(\mathrm{log}(3))\end{array}$

By the Mean Value Theorem and since $\frac{1}{n\mathrm{log}(n)}$ is decreasing, for some $\kappa \in (k-1,k)$

$\begin{array}{}\text{(3)}& \frac{1}{k\mathrm{log}(k)}\le \frac{1}{\kappa \mathrm{log}(\kappa )}=\mathrm{log}(\mathrm{log}(k))-\mathrm{log}(\mathrm{log}(k-1))\end{array}$

Therefore, the red difference in $(2)$ is decreasing in $n$ since

$\begin{array}{rl}\phantom{\rule{-1cm}{0ex}}& {\sum _{k=2}^{n}\frac{1}{k\mathrm{log}(k)}-\mathrm{log}(\mathrm{log}(n))}\\ \text{(4)}& & =\frac{1}{2\mathrm{log}(2)}-\mathrm{log}(\mathrm{log}(2))+\sum _{k=3}^{n}\left({\frac{1}{k\mathrm{log}(k)}-[\mathrm{log}(\mathrm{log}(k))-\mathrm{log}(\mathrm{log}(k-1))]}\right)\end{array}$

and by $(3)$, each blue term in $(4)$ is negative.

So, $(2)$ says that the red difference is decreasing and bounded below by $-\mathrm{log}(\mathrm{log}(2))$. Thus, the limit of the red difference exists and is at least $-\mathrm{log}(\mathrm{log}(2))\doteq 0.366512920581664$

Furthermore, $(2)$ also says that for each $n\ge 3$, the red difference is at most $\frac{1}{2\mathrm{log}(2)}+\frac{1}{3\mathrm{log}(3)}-\mathrm{log}(\mathrm{log}(3))\doteq 0.930712768370062$

Jaxon Hamilton

Expert

2022-07-17Added 3 answers

The function $f:x\mapsto \frac{1}{x\mathrm{log}x}$ is continuous non negative decreasing on $[2,+\mathrm{\infty})$ hence the sequence

${S}_{n}=\sum _{k=3}^{n}(\frac{1}{k\mathrm{log}k}-{\int}_{k-1}^{k}\frac{dx}{x\mathrm{log}x})$

is convergent, in fact

$f(n)-f(2)=\sum _{k=3}^{n}(f(k)-f(k-1))\le {S}_{n}\le 0$

Can you take it from here?

${S}_{n}=\sum _{k=3}^{n}(\frac{1}{k\mathrm{log}k}-{\int}_{k-1}^{k}\frac{dx}{x\mathrm{log}x})$

is convergent, in fact

$f(n)-f(2)=\sum _{k=3}^{n}(f(k)-f(k-1))\le {S}_{n}\le 0$

Can you take it from here?

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