$\sum_{2n\le x}\phi(2n)$

Question

We know that, denoting with $\phi$ the Euler's function we have $$ \sum_{n\le x}\phi(n) \sim \frac{3}{\pi^2}x^2. $$

What is the same asymptotic formula for $$ \sum_{2n\le x}\phi(2n)\,\,\,? $$

Answer 1

This is a bit of a fudge and would benefit from some tidying up!

Let $$ S(x) = \sum_{n\space odd, n\le x}\phi(n) $$ Then $$ \sum_{n\le x}\phi(n) = S(x) + \sum_{2||n, n\le x}\phi(n) + \sum_{4||n, n\le x}\phi(n) + ... $$ where the sums terminate with the largest power of $2$ that is less than $x$.

This is simple to rewrite in terms of $S$ $$ \sum_{n\le x}\phi(n) = S(x) + S({x\over2}) + 2S({x\over4}) + 4S({x\over8}) +... $$ Now the fudge. Let's assume $S(x)$ is asymptotically quadratic, i.e. $$ S(x) = kx^2 + o(x^2) $$ Substituting we obtain $$ {3\over k \pi^2} = 1 + {1\over 4} + {1\over 8} + {1\over 16} + ... = {3\over 2} $$ So $k = 2/\pi^2$ and $$ S(x) = {2\over \pi^2}x^2 + o(x^2) $$ To complete, $$ \sum_{2n\le x} \phi(2n) = \sum_{n\le x} \phi(n) - S(x) = {x^2 \over \pi^2} + o(x^2) $$

Okay, I did some pretty bad things here. I don't feel too bad about the convergence of the geometric series - even though I should have been more careful it's fairly obvious that any residue will be $o(x^2)$. The really bad bit is assuming the quadratic asymptote. If anyone can justify that please add to the post.