Limit and number theory

Let $\varphi $ denote the Euler function. Evaluate the limit:

$$\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k \leq n} \varphi(k)$$

Solution



We have that $\displaystyle \frac{1}{n^2} \sum_{k=1}^{n}\varphi(k)= \frac{1}{2n^2} \sum_{s=1}^{n}\sum_{t=1}^{n}I \left [ (s,t)=1 \right ] $ where $I(A)=1$ if $A$ is true and $0$ if $A$ is false.

To evaluate the second sum we use the inclusion-exclusion principal, hence:

$$\frac{1}{n^2} \sum_{s=1}^n \sum_{t=1}^n I[(s,t)=1] = \frac{1}{n^2}\left(n^2 - \sum_{p \leqslant n} \lfloor n/p \rfloor^2 + \sum_{p_1 < p_2 \leqslant n} \lfloor n/p_1p_2 \rfloor^2 - \cdots \right) =\\= \frac{1}{n^2} \sum_{k=1}^n \mu(k) \lfloor n/k \rfloor^2$$

Hence we now easily check that:

$$\lim_{n \to +\infty} \frac{1}{n^2} \sum_{k \leqslant n} \phi(k) = \frac{1}{2} \sum_{k=1}^{\infty} \frac{\mu(k)}{k^2}$$

The last sum is the inverse $\zeta$ function and converges to $\displaystyle \frac{6}{\pi^2}$. Hence the original limit is equal to $\displaystyle \frac{3}{\pi^2}$ and the exercise is complete. 

$(*) \;\;\;\; $ $n=p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ is the analysis of $n$ in prime factors.