1
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$$\lim_{n\to \infty}\frac{\frac{1}{2\ln2}+\frac{1}{3\ln3}+\ldots+\frac{1}{n\,\ln\,n}}{\ln(\ln\,n)}$$

The result is $1$ (according to the book, though it does not show the steps, which I'm interested in).

I've applied the theorem and it led me to an equally unhelpful limit.

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    Note that $\log(\log(n+1)) - \log(\log(n)) = \log\left(\frac{\log(n+1)}{\log(n)}\right) = \log\left(1 + \frac{\log(1 + 1/n)}{\log(n)}\right) \sim \frac{1}{n\log(n)}$ by using $\log(1+x)\sim x$ twice. – 2017-01-26
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    Cesaro-Stolz is one good way. A sort of the opposite approach is to approximate the sum by an integral, which you can explicitly compute. – 2017-01-26
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    So, I apply the theorem: $\lim_{n\to \infty} \frac{\frac{1}{(n+1)log(n+1)}-\frac{1}{nlogn}}{log(\frac{log(n+1)}{logn})}$ and following the method of @Winther , I get to this : $\lim_{n\to \infty} \frac{\frac{1}{(n+1)log(n+1)}-\frac{1}{nlogn}}{\frac{1}{nlogn}}$. Now, this becomes : $\lim_{n\to \infty} \frac{\frac{1}{(n+1)log(n+1)}}{\frac{1}{nlogn}}-\lim_{n\to \infty} \frac{\frac{1}{nlogn}}{\frac{1}{nlogn}}$. The second limit is equal to 1, meanwhile the first is $\lim_{n\to \infty} \frac{nlogn}{(n+1)log(n+1)}$ = 1 ? – 2017-01-26
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    You have done a small mistake in applying the theorem. What is your $a_n$? Note that if $a_n = \sum_{k=2}^n x_k$ then $a_{n+1} - a_n = x_{n+1}$. If you just fix this mistake then you won't get that last term and you would get the correct result $1$. – 2017-01-26
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    Which book is this with such great questions ? – 2017-01-26
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    Oh, yes, yes indeed. Thank you, @Winther. Now I understand, my $x_{n+1}$ should be $\frac{1}{(n+1)ln(n+1)}$. Should I make any further edits to this, or will it be clear to people searching for the same problem? – 2017-01-26
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    No need to edit the question (as you only mentioned this in the comments). Someone will likely post an answer and if not then since you already have solved it yourself feel free to answer your own question. – 2017-01-26
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    @ReneSchipperus Here it is: http://www.edumanager.ro/community/documente/concursuri_internationale_vol_2.pdf , you can find the problem on page 16. The book is in Romanian, though. – 2017-01-26
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    Thanks just out of curiosity what does $\overline{\mathbb{R}}$ mean in this book ? – 2017-01-26
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    @ReneSchipperus $\overline{\mathbb{R}}$ stands for the $[-\infty;\infty]$ interval, so $\mathbb{R}$ including $-\infty$ and $\infty$. I know it in the context of limits, for example $l=lim_{n\to \infty}2n+1=\infty$, where l $\in$ $\overline{\mathbb{R}}$. – 2017-01-26

2 Answers 2

2

If you apply Stolz you get

$$\frac{1}{n\ln n(\ln \ln n-\ln\ln(n-1))}$$ Now take $e$ to the power of the denominator you get $$\left(\frac{\ln n}{\ln(n-1)}\right)^{n\ln n}$$ Write this as $$\left(1+\frac{a_n}{n\ln n}\right)^{n\ln n}$$

Where $$a_n=n\ln n \frac{\ln (1+\frac{1}{n-1})}{\ln (n-1)}$$ its easy to see that $a_n\to 1$. And thus $$\left(1+\frac{a_n}{n\ln n}\right)^{n\ln n}\to e$$

Thus the denominator limits to $\ln e=1$ so the original expression above limits to $1$.

2

We apply the Stolz-Cesaro theorem: $$lim_{n\to \infty}\frac{(\frac{1}{2ln2}+\frac{1}{3ln3}+...+\frac{1}{n\,ln\,n}+\frac{1}{(n+1)\,ln(n+1)})-(\frac{1}{2ln2}+\frac{1}{3ln3}+...+\frac{1}{n\,ln\,n})}{ln(ln(n+1))-ln(ln\,n)}=lim_{n\to \infty}\frac{\,\frac{1}{(n+1)\,ln(n+1)}\,}{\,ln(\frac{ln(n+1)}{ln\,n})\,}$$

We know that: $$ln\left(\frac{ln(n+1)}{ln\,n}\right)=ln\left(\frac{ln(n(1+\frac{1}{n}))}{ln\,n}\right)=ln\left(\frac{ln\,n+ln(1+\frac{1}{n})}{ln\,n}\right)=ln\left(1+\frac{ln(1+\frac{1}{n})}{ln\,n}\right)$$

Now, $log(1+x)\ \sim x$, so $ln\left(1+\frac{ln(1+\frac{1}{n})}{ln\,n}\right) \sim \frac{ln(1+\frac{1}{n})}{ln\,n} \sim \frac{\frac{1}{n}}{ln\,n}=\frac{1}{n\,ln\,n}$ .

( "$\sim$" means "approximately equal to" in this context)

So: $$lim_{n\to \infty}\frac{\,\frac{1}{(n+1)\,ln(n+1)}\,}{\,ln(\frac{ln(n+1)}{ln\,n})\,}=lim_{n\to \infty}\frac{\,\frac{1}{(n+1)\,ln(n+1)}\,}{\,\frac{1}{n\,ln\,n}\,}=lim_{n\to \infty}\frac{n\,ln\,n}{(n+1)\,ln(n+1)}$$ which, in turn, is equal to $$\lim_{n\to \infty} \frac{n}{n+1}\cdot\lim_{n\to \infty}\frac{ln\,n}{ln(n+1)}=1\cdot1=1$$

So 1 is our final answer.