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! $$ Show\ \exists \ \beta \in [0,1], so\ that ..\\ \lim_{n\rightarrow \infty} \left(\sum^{n}_{k=2} \dfrac{1}{k \cdot \log(k)} - \log(\log(n))\right) = \beta\ \\ $$ 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.
On
Use the following lemma:
Let $f(x)$ be a decreasing positive function with $\lim_{x \to \infty} f(x)=0$
Then the sequence $X_n=\int_1^nf(t)dt -\sum_{i=1}^nf(i)$ converges.
In order to prove it, use the following facts:
1.$\int_1^nf(t)dt=\sum_{i=2}^n\int_{i-1}^if(t)dt$
2.$f(i)< \int_{i-1}^if(t)dt<f(i-1)$
Now just use the alternating series (Leibniz) test. (It might be helpful to construct and use the following series $a_{2i+1}=f(i)$ and $a_{2i}=\int_{i}^{i+1}f(t)dt$)
On
The function $f:x\mapsto\frac1{x\log x}$ is continuous non negative decreasing on $[2,+\infty)$ hence the sequence $$S_n=\sum_{k=3}^n\left(\frac1{k\log k}-\int_{k-1}^k \frac{dx}{x\log x}\right)$$ is convergent, in fact $$f(n)-f(2)=\sum_{k=3}^n(f(k)-f(k-1))\le S_n\le0$$
Can you take it from here?
Comparing sum and integral, we get $$ \begin{align} \frac1{n\log(n)}+\int_2^n\frac{\mathrm{d}x}{x\log(x)} &\le\sum_{k=2}^n\frac1{k\log(k)}\\ &\le\frac1{2\log(2)}+\frac1{3\log(3)}+\int_3^n\frac{\mathrm{d}x}{x\log(x)}\tag{1} \end{align} $$ Thus, $$ \begin{align} \frac1{n\log(n)}-\log(\log(2)) &\le\color{#C00000}{\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))}\\ &\le\frac1{2\log(2)}+\frac1{3\log(3)}-\log(\log(3))\tag{2} \end{align} $$ By the Mean Value Theorem and since $\frac1{n\log(n)}$ is decreasing, for some $\kappa\in(k-1,k)$, $$ \frac1{k\log(k)}\le\frac1{\kappa\log(\kappa)}=\log(\log(k))-\log(\log(k-1))\tag{3} $$ Therefore, the red difference in $(2)$ is decreasing in $n$ since $$ \begin{align} \hspace{-1cm}&\color{#C00000}{\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))}\\ &=\frac1{2\log(2)}-\log(\log(2))+\sum_{k=3}^n\left(\color{#0000FF}{\frac1{k\log(k)}-[\log(\log(k))-\log(\log(k-1))]}\right)\tag{4} \end{align} $$ and by $(3)$, each blue term in $(4)$ is negative.
So, $(2)$ says that the red difference is decreasing and bounded below by $-\log(\log(2))$. Thus, the limit of the red difference exists and is at least $-\log(\log(2))\doteq0.366512920581664$.
Furthermore, $(2)$ also says that for each $n\ge3$, the red difference is at most $\frac1{2\log(2)}+\frac1{3\log(3)}-\log(\log(3))\doteq0.930712768370062$.