An elementary proof for a limit? [closed]

Question

This question is motivated by pedagogical reason, not research. I will provide a simple proof for contrast, but I would like to see another approach that does not involve integrals, instead even more elementary tools.

Prove that the sequence $a_n$ converges if $$a_n=1+\sum_{k=2}^n\frac1{k\log k}-\log\log n.$$ Proof. Rewrite the given sequence as follows $$a_n=1+\sum_{k=2}^n\frac1{k\log k}-\log\log(n+1)+\log\log(n+1)-\log\log n.$$ Since $c_n:=\log\log(n+1)-\log\log n=\log\left(1+\frac{\log(1+1/n)}{\log n}\right)\rightarrow0$, as $n\rightarrow\infty$, we know $a_n$ converges iff $b_n$ converges; where \begin{align*} b_n:&=1+\sum_{k=2}^n\frac1{k\log k}-\log\log(n+1) \\ &=1-\log\log 2+\sum_{k=2}^n\frac1{k\log k}-\int_2^{n+1}\frac{dx}{x\log x}. \end{align*} This allows to compare the integral $\int_2^{n+1}\frac{dx}{x\log x}$ which is dominated by the Upper Riemann sum $\sum_{k=2}^n\frac1{k\log k}$. Therefore, we have positivity of $$d_n:=\sum_{k=2}^n\frac1{k\log k}-\int_2^{n+1}\frac{dx}{x\log x}>0$$ as well as monotonicity $d_{n+1}>d_n$. This becomes more apparent if you draw a picture. Next, notice that $d_n$ represents the excess area between the area under $f(x)=\frac1{x\log x}$ and the Riemann rectangle. Let's estimate $d_n$ by the difference between the Upper and Lower Riemann sums to compute rectangular areas (remember: the line segment $[2,n+1]$ is partitioned throughout by unit segments). Anyways, we obtain $$d_n<\sum_{k=2}^n\left(\frac1{k\log k}-\frac1{(k+1)\log(k+1)}\right)=\frac1{2\log 2}-\frac1{(n+1)\log(n+1)}<\frac1{2\log2}$$ which illustrates boundedness of the sequence $d_n$ and hence that of $b_n$. We know that any increasing sequence bounded from above is convergent. We conclude $b_n$ (and thus $a_n$) is convergent. The proof is complete. $\square$

Answer 1

As usually, you may replace integration by Lagrange mean value theorem. We have, denoting $f(x)=\log\log x$, $$ a_{n-1}-a_{n}=(f(n)-f(n-1))-\frac1{n\log n}=f'(\theta_n)-\frac1{n\log n}=\frac1{\theta_n\log \theta_n}-\frac1{n\log n},\\ n-1\leqslant \theta_n\leqslant n. $$ So, $a_{n-1}-a_{n}$ is positive, but the series $\sum (a_{n-1}-a_n)$ is dominated by a telescopic series $\sum (f'(n-1)-f'(n))$, this implies that the series $\sum (a_{n-1}-a_n)$ converges, it is equivalent to the fact that $a_n$ converges.

If you try to avoid also derivatives, I should ask what at all you know about logarithms. If you know somehow, say, that $\log t<t-1$ for $t>0$, you may rewrite this inequality as $$\frac1x< \frac{\log x-\log y}{x-y}<\frac1y$$ for $x>y>0$ (for $t=x/y$, $t=y/x$), hence $$ \frac{\log\log x-\log\log y}{x-y}=\frac{\log\log x-\log\log y}{\log x-\log y}\cdot \frac{\log x-\log y}{x-y} $$ belongs to the interval $(\frac1{x\log x},\frac1{y\log y})$ for $x>y>1$. This is what we really use applying Lagrange theorem in the proof above.

Answer 2

Adapting Fedor's argument, we have

$$ a_n-a_{n-1}=\frac{1}{n\log(n)}+\log\log(n-1)-\log\log n= \frac{1}{n\log(n)}+\log\left(\frac{\log(n-1)}{\log n}\right), $$ and, similarly to the calculation from your post, $$ \log\left(\frac{\log(n-1)}{\log n}\right)=\log\left(1+\frac{\log\frac{n-1}{n}}{\log n}\right)=\log\left(1+\frac{-\frac{1}{n}+O\left(\frac{1}{n^2}\right)}{\log n}\right). $$ Finally, $$ \log\left(1+\frac{-\frac{1}{n}+O\left(\frac{1}{n^2}\right)}{\log n}\right)= \frac{-\frac{1}{n}+O\left(\frac{1}{n^2}\right)}{\log n}+O\left(\frac{1}{n^2}\right)=-\frac{1}{n\log(n)}+O\left(\frac{1}{n^2}\right). $$ It follows that $a_n-a_{n-1}=O\left(\frac{1}{n^2}\right)$, and so $a_n$ converges.

The only thing used here is $\log(1+x)=x+O(x^2)$ for, say, $|x|<1/2$. That can be easily established using any reasonable rigorous definition of logarithm.