Limit and Euler’s function

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

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

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 $I(A)=0$ if $A$ is false.

We are invoking the exclusion – inclusion principal to evaluate the second sum. Thus:

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

and thus we conclude that

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

The latter sum is known to converge to $\frac{6}{\pi^2}$ and thus the original limit is equal to

$\displaystyle \ell = \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k \leq n} \varphi(k) = \frac{3}{\pi^2}$

Limit and Euler’s function

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