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I am trying to calculate the following limit:

$$ \lim_{n\to\infty} \dfrac{\log\left( \dfrac{1}{\log(n+1)}-\dfrac{1}{\log(n+2)} \right)}{\log(n^{-2})} $$ Doing some numerical experiments my guess for the limit is $1/2$. My problem is with the expression of the numerator. I have got the lower bound $$ \dfrac{1}{\log(n+1)}-\dfrac{1}{\log(n+2)} \geq \dfrac{C}{n(\log n)^2} $$ where $C>0$ is a constant. This bound allows me to note that my limit is bounded below by $$ \lim_{n\to\infty}\dfrac{\log\left( \dfrac{C}{n(\log n)^2}\right)}{\log n^{-2}}=\dfrac{1}{2}+C'\cdot \lim_{n\to\infty}\dfrac{\log\log n}{\log n}=\dfrac{1}{2} $$ so $1/2$ is indeed a lower bound for my limit. Now, when I try to get an upper bound, usually my bounds for the logarithm are too strong and do not provide me any useful information for the limit. Any help would be appreciated.

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    I think your approach is correct and the answer is $1/2$. Rather using approximations/bounds just directly manipulate the given expression and obtain the final answer. – 2017-01-04

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We can proceed as follows \begin{align} L &= \lim_{n \to \infty}\dfrac{\log\left(\dfrac{1}{\log(n + 1)} - \dfrac{1}{\log(n + 2)}\right)}{\log n^{-2}}\notag\\ &= -\frac{1}{2}\lim_{n \to \infty}\dfrac{\log\log\left(\dfrac{n + 2}{n + 1}\right) - \log\log(n + 1) - \log\log(n + 2)}{\log n}\notag\\ &= - \frac{1}{2}\left\{\lim_{n \to \infty}\dfrac{\log\log\left(\dfrac{n + 2}{n + 1}\right)}{\log n} - \frac{\log\log(n + 1)}{\log n} - \frac{\log\log(n + 2)}{\log n}\right\}\notag\\ &= - \frac{1}{2}\lim_{n \to \infty}\dfrac{\log\log\left(\dfrac{n + 2}{n + 1}\right)}{\log n}\notag\\ &= -\frac{1}{2}\lim_{n \to \infty}\dfrac{\log\left((n + 1)\log\left(1 + \dfrac{1}{n + 1}\right)\right) - \log(n + 1)}{\log n}\notag\\ &= \frac{1}{2}\lim_{n \to \infty}\frac{\log(n + 1)}{\log n}\notag\\ &= \frac{1}{2}\notag \end{align}

Here we have used the limits $$\lim_{x \to \infty}\frac{\log x}{x} = 0, \lim_{x \to 0}\frac{\log(1 + x)}{x}=1$$ Thus putting $x = 1/(n + 1)$ in second limit above we see that $(n + 1)\log\left(1 + \dfrac{1}{n + 1}\right) \to 1$. Also note that $$\frac{\log(n + 1)}{\log n} = 1 + \frac{n\log(1 + (1/n))}{n\log n}\to 1$$ and similarly $\log(n + 2)/\log n \to 1$. Therefore $$\frac{\log\log(n + 1)}{\log n} = \frac{\log\log(n + 1)}{\log(n + 1)}\cdot\frac{\log(n + 1)}{\log n} \to 0\cdot 1 = 0$$ and the same logic applies to $\log\log(n + 2)/\log n$.