A limit with arctan function and Euler $\varphi$ function The following limit has stumbed upon a while. I don't have the experience to work with these kind of limit as well as these special functions like $\varphi$.
Prove that
$$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\varphi(k) \arctan \left(\frac{k}{n} \right)}{k (n+k)} = \frac{3 \log 2}{4\pi}$$
where $\varphi$ is the Euler's totient function. How would one go to prove it? 
 A: We can use the fact that
$$
\frac{\phi(k)}k = \sum_{d\mid k} \frac{\mu(d)}d,
$$
where $\mu$ is the Möbius function and the sum is over all positive divisors $d$ of $k$. Then
\begin{align*}
\sum_{k=1}^{n} \frac{\varphi(k) \arctan \left(\frac{k}{n} \right)}{k (n+k)} &= \sum_{k=1}^{n} \frac{\arctan \left(\frac{k}{n} \right)}{n+k} \sum_{d\mid n} \frac{\mu(d)}d \\
&= \sum_{d\le n} \frac{\mu(d)}d \sum_{\substack{k\le n \\ d\mid k}} \frac{\arctan \left(\frac{k}{n} \right)}{n+k} \\
&= \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}},
\end{align*}
where we used the change of variables $k=jd$ in the last equality. For each fixed $d$, the inner sum is a Riemann sum for $\int_0^1 \frac{\arctan x}{1+x}\,dx$; since this integrand is bounded and $\sum_{d=1}^\infty \frac{\mu(d)}{d^2}$ converges absolutely to $\frac6{\pi^2}$, the dominated convergence theorem implies that
\begin{align*}
\lim_{n\to\infty} \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} &= \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \lim_{n\to\infty} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} \\
&= \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \int_0^1 \frac{\arctan x}{1+x}\,dx \\
&= \frac6{\pi^2} \int_0^1 \frac{\arctan x}{1+x}\,dx,
\end{align*}
This reduces the problem to the evaluation $\int_0^1 \frac{\arctan x}{1+x}\,dx = \frac{\pi\log 2}8$, which is itself nontrivial but which has been answered here before (modulo an integration by parts which is natural to try).
