$\sum_{2n\le x}\phi(2n)$ 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
$$
\sum_{2n\le x}\phi(2n)\,\,\,?
$$
 A: 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.
