Prove $\lim_{n\to\infty}\phi(n)=\infty$.

Question

How can I show $\displaystyle\lim_{n\to\infty}\phi(n)=\infty$?

Obviously as $n$ approaches infinity it will have infinitely many prime factorizations, so the product series approaches infinity as well. How can I formally prove this?

Answer 1

The cheapest way is $$ \phi(n) \geq \sqrt {\frac{n}{2}}, $$ with equality at $2$ only.

There is a slightly weaker result that is sometimes given as a question here, maybe something along the lines of $(1/2) \sqrt n,$ which does not require the full procedure... anyway, see Is the Euler phi function bounded below? where the setting is Ramanujan's, the same proofs as for Colossally Abundant Numbers and Superior Highly Composite Numbers. We get an infinite sequence of inequalities, where the overall shape is the Rosser and Schoenfeld thing, but these are easier to calculate and better for small $n.$ Anyway, after the square root, the next sensible one is $$ \phi(n) \geq 2 \left( \frac{n}{6} \right)^{2/3}, $$ with equality at $6$ only.

Answer 2

You need to prove that there are finitely many $n$ such that $\phi$(n)=$N$ for $N$ a positive integer. To prove this, you need to prove that the set of solutions $n$ to this equation is a finite set. First, prove that there are only finitely many primes possible in the prime factorization of $n$, and second that there are finitely many possible exponents in n=$(p_1)^a(p_2)^b(p_3)^c....$

To prove that there are finitely many primes in the prime factorization, we need to prove that n cannot have a prime factor greater than $N+1$, for if it does have one, we may choose the smallest such prime $p$ such that $q$$\geq$$p$ and $n=q^k(r)$ where $r$ is a positive integer relatively prime to $q$.

So now $\phi$(n) = $\phi(q^k)$$\phi(r)$ because $\phi$ is a multiplicative function. But now we get $\phi(n)$=$q^{k-1}(q-1)\phi(r)$ $\geq$ $(q-1)$ > N. But $\phi(n)$=N, so this is a contradiciton.

Now to prove that there are finitely many exponents, consider the prime factorization of n as $p_1$, $p_2$, .... $p_m$. There can only be finitely many exponents in the original representation of $n$ above because $\phi(n)$ would contain the term $(p_i)^{s-1}(s-1)$ which is larger than $N$ for sufficiently large $s$.

Now that we have proven that there are finitely many solutions n to the equation $\phi(n)=N$, we can prove the limit in question.

There are only finitely many n such that $M$ $\geq$ $\phi(n)$, we may choose the largest such n, then let $m$ > $n$ where $m$ is a positive integer. It is clear that $\phi(m)$ $\geq$ $M$, so the limit as $n$ tends to infinity of $\phi(n)$ is infinity.

Answer 3

You can use part of Theorem 15 of Rosser and Shoenfeld's paper http://www.math.wvu.edu/~mays/745/Rosser-Schoenfeld%20(euclid.ijm.1255631807).pdf which states that $$ \frac{n}{\phi(n)} < e^c \log \log n + \frac{5}{2 \log\log n} $$ where $c=0.57721566...$ is Euler's constant, for all $n>223092870$.

Added: Or, see Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, page 84, Theorem 4, wherein he proves that $\phi(n)$ has minimal order $$ e^{-\gamma} \frac{ n } {\log \log n}$$ where $\gamma$ is Euler's constant.

Added #2: More basically, we have $$ \phi(n) = n \prod_{p^{\alpha} || n} \left( 1-\frac{1}{p} \right) \ge n \left(1-\frac{1}{2} \right)\left(1-\frac{1}{3}\right)^{\omega(n)-1} \\ \ge n \left( \frac{1}{2} \right) \left( \frac{2}{3} \right)^{\frac{\log n}{\log 2} -1 } \\ = \frac{3}{4} n ^{2 - (\log 3)/(\log 2)} $$ and the last expression tends to $\infty$ as $n$ does.