Euler phi function: $\sum_{n=1}^{N}\sum_{d\mid n}d\cdot\phi(d)$ I am trying to find the value of $$\sum_{n=1}^{N}\sum_{d|n}d*\phi(d)$$  Is there a method to evaluate this for large N? $\phi(d)$ is the Euler phi function.
 A: This one can be treated using either the 
Wiener-Ikehara theorem, which
gives  the  dominant  asymptotics,  or Mellin-Perron,  which  predicts
additional  terms  of  the  expansion.  In either  case  we  need  the
appropriate Dirichlet series.
To find this series recall first that
$$\sum_{d|n} \varphi(d) = n$$
which implies that
$$\sum_{q\ge 1} \frac{\varphi(q)}{q^s} = 
\frac{\zeta(s-1)}{\zeta(s)}.$$
Introduce $$a(n) = \sum_{d|n} d \times \varphi(d)
= n \sum_{d|n} \varphi(d) \times \left(\frac{d}{n}\right)
= n \sum_{d|n} \varphi(d) \times \left(\frac{n}{d}\right)^{-1}.$$
This yields that
$$\sum_{q\ge 1} \frac{a(n)/n}{n^s}
= \frac{\zeta(s-1)}{\zeta(s)} \zeta(s+1).$$
which finally produces
$$\sum_{q\ge 1} \frac{a(n)}{n^s}
= \frac{\zeta(s-2)}{\zeta(s-1)} \zeta(s).$$
Call this $L(s).$ The dominant pole of $L(s)$ is at $s=3.$
By the Wiener-Ikehara theorem we thus have
$$\sum_{q\le N} a(q) \sim
\frac{\zeta(3)}{\zeta(2)} \frac{N^3}{3}
= \frac{6}{\pi^2} \zeta(3) \frac{N^3}{3}.$$
Using  Mellin-Perron we  can predict  another term.  The Mellin-Perron
integral is
$$\frac{1}{2\pi i}
\int_{7/2-i\infty}^{7/2+i\infty} L(s)\frac{N^s}{s} ds.$$
Shifting this integral to the left we obtain the term for the pole at $s=3$ plus an additional term from
the pole at $s=1,$ which is
$$\frac{-1/12}{-1/2} \frac{N}{1} = \frac{1}{6} N$$
for our final formula
$$\sum_{q=1}^N a(q) \sim
\frac{2}{\pi^2} \zeta(3) N^3
+ \frac{1}{6} N
+ \frac{1}{2} \sum_{d|N} d \times \varphi(d).$$
The relative error in percent  of this approximation is $0.186\%$ for
$N=20$. It is $0.0275\%$ for $N=2000$ and $0.000128\%$ for $N=20000.$
Note. What we have not included here is the contribution from the non-trivial zeros of the Riemann zeta function, which (assuming they are simple) is
$$\sum_{\rho} \frac{1}{\zeta'(\rho)} \times
\zeta(-1+\rho) \zeta(1+\rho) \frac{N^{1+\rho}}{1+\rho}.$$
A: Another, more elementary derivation of the same result is given here:
J. Sándor and A. V. Kramer. Über eine Zahlentheoretische Funktion. Mathematica Moravica,
3:53–62, 1999. URL: http://www.moravica.ftn.kg.ac.rs/Vol_3/10-Sandor-Kramer.pdf.
Here is a comment to the previous answer, but I don't have enough "reputation" to write a comment:
The asymptotic "final formula" of Marko Riedel's answer does not include an explicit error term, and I did not follow the derivation, but in any case, the linear term N/6 does not make sense to me. The summand $a(N)$ of the series fluctuates in a quadratic range. For example, when $N$ is a prime, $a(N)=N(N-1)+1$; when $N$ is a power of 2, $a(N)=(2N^2+1)/3$.
Thus, when writing an approximation to $\sum_{q=1}^N a(q)$ [or, bringing the last term on the right side to the left as in the "final formula", to $\sum_{q=1}^{N-1} a(q) + 1/2\cdot a(N)$] by a smooth function like $CN^3$ or $C_1N^3+C_2N$, one has to expect a quadratic error term $O(n^2)$, which would swallow any linear or lower-order terms. (Otherwise, by subtracting the approximation for $N-1$ from the approximation for $N$, one could predict $a(N)$ [or $a(N)+a(N-1)$] to better than quadratic order.)
