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. $\square$


closed as off-topic by R W, Alexey Ustinov, Myshkin, Lucia, Marco Golla Oct 2 '16 at 8:13

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  • 2
    $\begingroup$ I would say that $\frac{1}{k \log k} - \int_k^{k+1} \frac{1}{x \log x}dx = \int_k^{k+1}(\frac{1}{k \log k} - \frac{1}{x \log x})dx = \int_k^{k+1} \int_k^x (\frac{-1}{t \log t})' dt dx$ $ = \int_k^{k+1} \int_k^x \frac{1+\log t}{t^2 \log^2 t} dt dx = \mathcal{O}(k^{-2})$ and I think it is more elementary than any calculus-less approach $\endgroup$ – reuns Oct 1 '16 at 23:08

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.

  • $\begingroup$ This is cool, Fedor. Yet, we're using a tool (derivatives) which is an equivalent for integrals (as you noted). Is it possible to stay away from both (sorry I was not very clear) and apply only ideas of sequences, such as Cauchy and basic limit theorems for sequences? $\endgroup$ – T. Amdeberhan Oct 1 '16 at 22:41
  • $\begingroup$ Yes, we may assume we know $\log t<t-1$ because iff $\log(1+t)<t$ iff $1+t<e^t$ iff $e^t>2^t=(1+1)^t>1+t$ from Binomial Theorem. $\endgroup$ – T. Amdeberhan Oct 2 '16 at 2:05
  • $\begingroup$ The last inequality fails for $t<1$. $\endgroup$ – Fedor Petrov Oct 2 '16 at 6:07
  • $\begingroup$ I need both $t=x/y, t=y/x$. And your $t$ is my $t-1$, even worse. $\endgroup$ – Fedor Petrov Oct 2 '16 at 13:01
  • $\begingroup$ You are right, Fedor. $\endgroup$ – T. Amdeberhan Oct 2 '16 at 13:03

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.

  • $\begingroup$ I'm not sure we can say $a_n-a_{n-1}=O(1/n^2)$ implies $a_n$ converges, in general. Take $a_n=\log(n^2)$ for $n$ even; $a_n=\log(1+n^2)$ for $n$ odd. Am I right? We need to follow through Fedor's argument on the series $\sum (a_{n-1}-a_n)$. $\endgroup$ – T. Amdeberhan Oct 2 '16 at 2:12
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    $\begingroup$ @T.Amdeberhan you are not right, this sequence does not satisfy $a_n-a_{n-1}=O(1/n^2)$, you maybe forgot that $n$ is different for $a_n$ and for $a_{n-1}$. Of course a series $\sum c_n$ converges if $c_n=O(1/n^2)$. $\endgroup$ – Fedor Petrov Oct 2 '16 at 6:14

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