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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.

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3 Answers 3

up vote 2 down vote accepted

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 $\frac1{2\log(2)}+\frac1{3\log(3)}-\log(\log(3))\doteq0.930712768370062$.

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Thank you very much!! i have two questions: (i) ok i know, if k' is in (k-1,k) than 1/k*log(k) <= 1/k'*log(k') but why is 1/k'*log(k') = log(log(k)-log(log(k-1)) equal? (ii) The second Series in (4). The Series starts at k=2, when it starts at k=2 "log(log(2))-log(log(2-1))" isn't defined. So, why the equality is valid? –  Landau Apr 5 at 23:39
    
@Landau: I have added a link to the Mean Value Theorem. That should answer your first question about $\kappa$. I have adjusted the limits in $(4)$. –  robjohn Apr 6 at 10:00
    
@Landau: I notice that you've unaccepted and then reaccepted this answer a couple of times in the last two days. If this was simply a mis-click, no sweat. However, if you are trying to upvote, you need to have at least 15 reputation first. –  robjohn Apr 11 at 2:10
    
@ robjohn - Im a newbie here, it was a mis-click. I totally agree your answer! And yes I tried to vote for your answer that its useful, then i note i need 15 repps. at first, thanks rob :) –  Landau Apr 11 at 15:57

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?

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Well done, as usual! –  amWhy Apr 6 at 13:30
    
After robjohns solution i understood yours! Fantastic how much information in such 2, 3 lines is hidden - I love mathematics :) Thanks! –  Landau Apr 10 at 22:15
    
You're welcome.@Landau –  Sami Ben Romdhane Apr 10 at 22:54

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 $a_{2i}=\int_{i}^{i+1}f(t)dt$)

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