Does 
$$\frac{1}{N^2}\sum \_{d=1}^N \log d \sum \_{n=1}^{N/d} \frac{\phi(n)}{\log (dn)},$$


converges or not when $N$ goes to infinity? 



Average involving the Euler phi function
The expression converges to $0$, even when $\phi(n)$ is replaced by the larger $n$. The contribution from $n\le\sqrt N/d$ can be given by ignoring the logarithm in the denominator:
\begin{align\*}
\frac1{N^2} \sum\_{d=1}^N \log d \sum\_{n=1}^{\sqrt N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum\_{d=1}^N \log d \sum\_{n=1}^{\sqrt N/d} n \\\
&\lesssim \frac1{N^2} \sum\_{d=1}^N \log d \cdot \frac12 \bigg( \frac{\sqrt N}d \bigg)^2 \\\
&= \frac1{2N} \sum\_{n=1}^\infty \frac{\log d}{d^2} = \frac{|\zeta'(2)|}{2N}.
\end{align\*}
(The exact value of the sum over $d$ isn't important - the fact that it converges is enough to show that the limit equals $0$ as $N\to\infty$.) The remaining contribution can be given by simplifying the logarithm in the denominator:
\begin{align\*}
\frac1{N^2} \sum\_{d=1}^N \log d \sum\_{n=\sqrt N/d}^{N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum\_{d=1}^N \log d \sum\_{n=1}^{N/d} \frac n{\log \sqrt N} \\\
&\lesssim \frac2{N^2\log N} \sum\_{d=1}^N \log d \cdot \frac12 \bigg( \frac Nd \bigg)^2 \\\
&= \frac1{\log N} \sum\_{n=1}^\infty \frac{\log d}{d^2} = \frac{|\zeta'(2)|}{\log N},
\end{align\*}
which also tends to $0$ as $N\to\infty$.


