Upper and lower bounds on the totient summatory function

In this post I use manipulations as in Ramanujan's proof of Bertrand's postulate to calculate explicit upper and lower bounds on the totient summatory function \( \Phi(x) = \sum_{n \le x} \phi(n) \). This shows how Ramanujan's proof works in a simpler situation.


Here we go. Because \( \phi \star 1 = n \), it follows that \( \sum_{n =1}^\infty \Phi(\frac{x}{n}) = \sum_{n \le x} n \equiv F(x) \). Because \( \Phi \) is always positive, the upper bound \( F(x) \ge \Phi(x) \) follows trivially. To obtain a lower bound, calculate

\( \begin{align}
F(x) - 2 F(\frac{x}{2} )  &= \Phi(x) + \Phi(\frac{x}{2}) + \Phi(\frac{x}{3}) + \cdots  -  2 \Phi(\frac{x}{2}) -2 \Phi(\frac{x}{4 }) -\cdots \\
 &= \Phi(x) - \Phi(\frac{x}{2}) + \Phi(\frac{x}{3}) - \cdots \\
\end{align}
\)
This is an alternating series, thus \( F(x) - 2 F(\frac{x}{2} )  \le \Phi(x) \). If for simplicity I do not make a distinction between \( x \) and \( \lfloor x \rfloor \), then \( F(x) = \frac{1}{2} x ( x +1 ) \) and \( F(x) - 2 F(\frac{x}{2} ) = \frac{x^2}{4} \). One has thus

$$ \frac{x^2}{4} \le \Phi(x) \le \frac{1}{2} x ( x +1 ) $$

I will now check these inequalities in Mathematica. Here is a graph of values of \( \Phi(x) \) (blue dots) and the upper and lower bounds (black lines) for \( x \) up to 50.

This looks good. I also check for \( x \) up to 10000.

Both values are positive; the inequalities thus seem to be valid.
Remarks

• The proof is not complete because I have not made a distinction between \( x \) and \( \lfloor x \rfloor \)
• It is known that \( \Phi(x) \sim \frac{3}{\pi^2} x^2 \), see here for example.
• The calculations were performed in Mathematica 10 Home Edition.