Proofs involving the totient function

This page provides proofs for identities involving the totient function φ(k) and the Möbius function μ(k).

Sum of integers relatively prime to and less than or equal to n

Claim:

$$\sum_{1\le k\le n \atop {\gcd(k,n)=1}} k = \frac{1}{2} \, \varphi(n) \, n \quad \text{ for integers } n \ge 2.$$ Since gcd (n,n) ≠ 1 and gcd (k,n) = gcd (n−k,n) for all integers k, a change of index yields

$$\sum_{1\le k\le n \atop {\gcd(k,n)=1}} k \,= \sum_{1\le k\le n \atop {\gcd(k,n)=1}} (n-k)$$. Therefore

$$2\sum_{1\le k\le n \atop {\gcd(k,n)=1}} k \,= \sum_{1\le k\le n \atop {\gcd(k,n)=1}} (k+n-k) = \varphi(n)\,n$$.

Proofs of totient identities involving the floor function

The proof of the identity

$$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor$$

is by mathematical induction on n. The base case is n = 1 and we see that the claim holds:

$$\varphi(1)/1 = 1 = \frac{\mu(1)}{1} \left\lfloor 1\right\rfloor.$$

For the induction step we need to prove that

$$\frac{\varphi(n+1)}{n+1} = \sum_{k=1}^n\frac{\mu(k)}{k} \left(\left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor\right) + \frac{\mu(n+1)}{n+1}.$$

The key observation is that

$$\left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor \; = \; \begin{cases} 1, & \mbox{if }k|(n+1) \\ 0, & \mbox{otherwise, } \end{cases}$$

so that the sum is

$$\sum_{k|n+1,\; k<n+1} \frac{\mu(k)}{k} + \frac{\mu(n+1)}{n+1} = \sum_{k|n+1} \frac{\mu(k)}{k}.$$

Now the fact that

$$\sum_{k|n+1} \frac{\mu(k)}{k} = \frac{\varphi(n+1)}{n+1}$$

is a basic totient identity. To see that it holds, let p1v1p2v2…pqvq be the prime factorization of n+1. Then

$$\frac{\varphi(n+1)}{n+1} = \prod_{l=1}^q \left( 1 - \frac{1}{p_l} \right) = \sum_{k|n+1} \frac{\mu(k)}{k}$$

by definition of μ(k). This concludes the proof.

An alternate proof proceeds by substituting $\frac{\varphi(k)}{k} = \sum_{d|k}\frac{\mu(d)}{d}$ directly into the left side of the identity, giving $\sum_{k=1}^n \sum_{d|k}\frac{\mu(d)}{d}.$

Now we ask how often the term $\begin{matrix}\frac{\mu(d)}{d}\end{matrix}$ occurs in the double sum. The answer is that it occurs for every multiple k of d, but there are precisely $\begin{matrix}\left\lfloor\frac{n}{d}\right\rfloor\end{matrix}$ such multiples, which means that the sum is

$$\sum_{d=1}^n\frac{\mu(d)}{d}\left\lfloor\frac{n}{d}\right\rfloor$$

as claimed.

The trick where zero values of $\begin{matrix} \left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor \end{matrix}$ are filtered out may also be used to prove the identity

$$\sum_{k=1}^n\varphi(k) = \frac{1}{2}\left(1+ \sum_{k=1}^n \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^2\right).$$

The base case is n = 1 and we have

$$\varphi(1) = 1 = \frac{1}{2} \left(1+ \mu(1) \left\lfloor\frac{1}{1}\right\rfloor^2\right)$$

and it holds. The induction step requires us to show that

$$\varphi(n+1) = \frac{1}{2} \sum_{k=1}^n \mu(k) \left( \left\lfloor\frac{n+1}{k}\right\rfloor^2 - \left\lfloor\frac{n}{k}\right\rfloor^2 \right) \; + \; \frac{1}{2} \; \mu(n+1) \; \left\lfloor\frac{n+1}{n+1}\right\rfloor^2 .$$

Next observe that

$$\left\lfloor\frac{n+1}{k}\right\rfloor^2 - \left\lfloor\frac{n}{k}\right\rfloor^2 \; = \; \begin{cases} 2\frac{n+1}{k} - 1, & \mbox{if }k|(n+1) \\ 0, & \mbox{otherwise.} \end{cases}$$

This gives the following for the sum

$$\frac{1}{2} \sum_{k|n+1, \; k<n+1} \mu(k) \left( 2\frac{n+1}{k} - 1 \right) \; + \; \frac{1}{2} \; \mu(n+1) = \frac{1}{2} \sum_{k|n+1} \mu(k) \left( 2 \; \frac{n+1}{k} - 1 \right).$$

Treating the two inner terms separately, we get

$$(n+1) \sum_{k|n+1} \frac{\mu(k)}{k} - \frac{1}{2} \sum_{k|n+1} \mu(k).$$

The first of these two is precisely φ(n+1) as we saw earlier, and the second is zero, by a basic property of the Möbius function (using the same factorization of n + 1 as above, we have $\begin{matrix} \sum_{k|n+1} \mu(k) = \prod_{l=1}^q (1-1) = 0 \end{matrix}$.) This concludes the proof.

This result may also be proved by inclusion-exclusion. Rewrite the identity as

$$-1 + 2\sum_{k=1}^n\varphi(k) = \sum_{k=1}^n \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^2.$$

Now we see that the left side counts the number of lattice points (a, b) in [1,n] × [1,n] where a and b are relatively prime to each other. Using the sets Ap where p is a prime less than or equal to n to denote the set of points where both coordinates are divisible by p we have

|⋃pAp| = ∑p|Ap| − ∑p < q|Ap∩Aq| + ∑p < q < r|Ap∩Aq∩Ar| − ⋯ ± |Ap∩ ⋯ ∩As|.

This formula counts the number of pairs where a and b are not relatively prime to each other. The cardinalities are as follows:

$$\left| A_p \right| = \left\lfloor \frac{n}{p} \right\rfloor^2, \; \left| A_p \cap A_q \right| = \left\lfloor \frac{n}{pq} \right\rfloor^2, \; \left| A_p \cap A_q \cap A_r \right| = \left\lfloor \frac{n}{pqr} \right\rfloor^2, \; \ldots$$

and the signs are $\begin{matrix}-\mu(pqr\cdots)\end{matrix}$, hence the number of points with relatively prime coordinates is

$\mu(1)\, n^2 \; + \; \sum_p \mu(p) \left\lfloor \frac{n}{p} \right\rfloor^2 \; + \; \sum_{p<q} \mu(p q) \left\lfloor \frac{n}{pq} \right\rfloor^2 \; + \; \sum_{p<q<r} \mu(p q r) \left\lfloor \frac{n}{pqr} \right\rfloor^2 \; + \; \cdots$

but this is precisely $\sum_{k=1}^n \mu(k) \left\lfloor \frac{n}{k} \right\rfloor^2$ and we have the claim.

Average order of the totient

We will use the last formula of the preceding section to prove the following result:

$$\frac{1}{n^2} \sum_{k=1}^n \varphi(k)= \frac{3}{\pi^2} + \mathcal{O}\left(\frac{\log n }{n}\right)$$

Using x − 1 < ⌊x⌋ ≤ x we have the upper bound

$$\frac{1}{2 n^2} \left(1+ \sum_{k=1}^n \mu(k)\frac{n^2}{k^2}\right) = \frac{1}{2 n^2} + \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2}$$

and the lower bound

$$\frac{1}{2 n^2} \left(1+ \sum_{k=1}^n \mu(k)\left(\frac{n^2}{k^2} - 2\frac{n}{k} + 1\right)\right)$$ which is

$$\frac{1}{2 n^2} + \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} - \frac{1}{n} \sum_{k=1}^n \frac{\mu(k)}{k} + \frac{1}{2 n^2} \sum_{k=1}^n \mu(k)$$

Working with the last two terms and using the asymptotic expansion of the nth harmonic number we have

$$- \frac{1}{n} \sum_{k=1}^n \frac{\mu(k)}{k} > - \frac{1}{n} \sum_{k=1}^n \frac{1}{k} = - \frac{1}{n} H_n > -\frac{1}{n} (\log n +1)$$ and

$$\frac{1}{2 n^2} \sum_{k=1}^n \mu(k) > - \frac{1}{2 n}.$$

Now we check the order of the terms in the upper and lower bound. The term $\begin{matrix}\sum_{k=1}^n \frac{\mu(k)}{k^2}\end{matrix}$ is 𝒪(1) by comparison with ζ(2), where ζ(s) is the Riemann zeta function. The next largest term is the logarithmic term from the lower bound.

So far we have shown that

$$\frac{1}{n^2} \sum_{k=1}^n \varphi(k)= \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} + \mathcal{O}\left(\frac{\log n }{n}\right)$$

It remains to evaluate $\begin{matrix}\sum_{k=1}^n \frac{\mu(k)}{k^2}\end{matrix}$ asymptotically, which we have seen converges. The Euler product for the Riemann zeta function is

$$\zeta(s) = \prod_p \left(1 - \frac{1}{p^s} \right)^{-1} \mbox{ for } \Re(s)>1.$$

Now it follows immediately from the definition of the Möbius function that

$$\frac{1}{\zeta(s)} = \prod_p \left(1 - \frac{1}{p^s} \right) = \sum_{n \ge 1} \frac{\mu(n)}{n^s}.$$

This means that

$$\frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} = \frac{1}{2} \frac{1}{\zeta(2)} + \mathcal{O}\left(\frac{1}{n}\right)$$ where the integral $\begin{matrix} \int_{n+1}^\infty \frac{1}{t^2} dt\end{matrix}$ was used to estimate $\begin{matrix} \sum_{k>n} \frac{\mu(k)}{k^2}\end{matrix}.$ But $\begin{matrix}\frac{1}{2} \frac{1}{\zeta(2)} = \frac{3}{\pi^2}\end{matrix}$ and we have established the claim.

Average order of φ(n)/n

The material of the preceding section, together with the identity

$$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor$$

also yields a proof that

$$\frac{1}{n} \sum_{k=1}^n \frac{\varphi(k)}{k} = \frac{6}{\pi^2} + \mathcal{O}\left(\frac{\log n }{n}\right).$$

Reasoning as before, we have the upper bound

$$\frac{1}{n} \sum_{k=1}^n\frac{\mu(k)}{k}\frac{n}{k} = \sum_{k=1}^n \frac{\mu(k)}{k^2}$$ and the lower bound

$$-\frac{1}{n} \sum_{k=1}^n\frac{\mu(k)}{k} + \sum_{k=1}^n \frac{\mu(k)}{k^2}.$$

Now apply the estimates from the preceding section to obtain the result.

Inequalities

We first show that

$$\lim \inf \frac{\varphi (n)}{n}=0 \mbox{ and } \lim \sup \frac{\varphi (n)}{n}=1.$$

The latter holds because when n is a power of a prime p, we have

$$\frac{\varphi (n)}{n} = 1 - \frac{1}{p},$$

which gets arbitrarily close to 1 for p large enough (and we can take p as large as we please since there are infinitely many primes).

To see the former, let nk be the product of the first k primes, call them p1, p2, ..., pk. Let

$$r_k = \frac{\varphi (n_k)}{n_k} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right).$$

Then

$$\frac{1}{r_k} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right)^{-1} > \sum_{m=1}^{p_k} \frac{1}{m} = H_{p_k},$$

a harmonic number. Hence, by the well-known bound Hn > log n, we have

$$\frac{1}{r_k} > \log p_k.$$

Since the logarithm is unbounded, taking k arbitrarily large ensures that rk achieves values arbitrarily close to zero.

We use the same factorization of n as in the first section to prove that

$$\frac {6 n^2} {\pi^2} < \varphi(n) \sigma(n) < n^2$$.

Note that

$$\varphi(n) \sigma(n) = n \prod_{l=1}^q \left(1 - \frac{1}{p_l}\right) \prod_{l=1}^q \frac{p_l^{v_l+1}-1}{p_l-1} = n \prod_{l=1}^q \frac{p_l-1}{p_l} \; \frac{p_l^{v_l+1}-1}{p_l-1}$$

which is

$$n \prod_{l=1}^q \left(p_l^{v_l}-\frac{1}{p_l}\right) = n^2 \prod_{l=1}^q \left(1 - \frac{1}{p_l^{v_l+1}}\right).$$

The upper bound follows immediately since

$$\prod_{l=1}^q \left(1 - \frac{1}{p_l^{v_l+1}}\right) < 1.$$

We come arbitrarily close to this bound when n is prime. For the lower bound, note that

$$\prod_{l=1}^q \left(1 - \frac{1}{p_l^{v_l+1}}\right) \ge \prod_{l=1}^q \left(1 - \frac{1}{p_l^2}\right) > \prod_p \left(1 - \frac{1}{p^2}\right),$$

where the product is over all primes. We have already seen this product, as in

$$\prod_p \left(1 - \frac{1}{p^s}\right) = \sum_{n\ge 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$$

so that

$$\prod_p \left(1 - \frac{1}{p^2}\right) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}$$

and we have the claim. The values of n that come closest to this bound are products of the first k primes.