Efficient way to compute $\sum_{i=1}^n \varphi(i) $ Given some upper bound $n$ is there an efficient way to calculate the following:
$$\sum_{i=1}^n \varphi(i) $$
I am aware that:
$$\sum_{i=1}^n \varphi(i) = \frac 12 \left( 1+\sum_{i=1}^n \mu(i) \left \lfloor \frac ni \right\rfloor ^2 \right) $$
Where:
$\varphi(x) $ is Euler's Totient
$\mu(x) $ is the Möbius function 
I'm wondering if there is a way to reduce the problem to simpler computations because my upper bound on will be very large, ie: $n \approx 10^{11} $. 
Neither $\varphi(x) $, nor $\mu(x) $, are efficient to compute for a large bound of $n$   
Naive algorithms will take an unacceptably long time to compute (days) or I would need would need a prohibitively expensive amount of RAM to store look-up tables.
 A: You can compute $\mu$ efficiently by sieving on intervals. With $n\approx10^{11},$ intervals of length $10^7$ should work well, taking only a few megabytes. Time needed is $O(n\log\log n)$ and space is proportional to $\sqrt n$ or so.
To improve on this, note that $\lfloor n/i\rfloor$ will be constant on large stretches and so that doesn't need to be computed very often. You can then use fast techniques to compute $M(k)$ for a small number of $k$: $n/2,\ n/3,\ n/4,\ \ldots.$ Time needed is roughly $n^{5/6}$ and space is a bit over $n^{1/3}$ using Deléglise & Rivat. You can use sieving to finish the rest in the same time, though it will need a bit more space ($\sqrt n$ would suffice, $n^{5/12}$ is probably achievable without trouble). Practically speaking you'll probably choose somewhat more memory to speed to computation.
Note that this is sequence A002088 in the OEIS.
Edit: I should also mention sequence A064018 which has
$$
\sum_{n=1}^{10^{11}}\varphi(n)=3039635509283386211140
$$
(as computed by the late Donovan Johnson) in addition to sums up to other powers of 10.
A: If you can efficiently compute $\sum_{i=1}^n \varphi(i)$, then you can efficiently compute two consecutive values, so you can efficiently compute $\varphi(n)$. Since you already know that $\varphi(n)$ isn't efficiently computable, it follows that $\sum_{i=1}^n \varphi(i)$ isn't either. 
