Can the existence of infinitely many even perfect numbers be settled by a diagonal argument? Say a (finite or not) sequence of strictly increasing positive integers $(u_{i})_{i\in I}$ is a 'Euclid sequence' if and only if the sum of reciprocals of all the $(u_i)$ equals $2$.
Now suppose we list all such sequences by increasing number of terms. The first sequence would be $(v_{1,i})=(1, 2, 3, 6)$, the second $(v_{2,i})=(1, 2, 3, 7, 42)$, the third $(v_{3,i})=(1, 2, 4, 7, 14, 28)$ and so on.
Consider now the sequence $(v_{i,i})=(1, 2, 4, 8,...)$. Is it a Euclid sequence?  
Many thanks in advance.

By numerical search I have found that the first few Euclid sequences are the following (ordered, per comment, first by length, second by sum of elements):
$$\begin{array}
&&1&2&3&6\\
&1&2&4&6&12\\
&1&2&3&10&15\\
&1&2&4&5&20\\
&1&2&3&9&18\\
&1&2&3&8&24\\
&1&2&3&7&42
\end{array}$$
and that these are all Euclid sequences up to 5 terms long.
We see from here that $(v_{i,i})$ starts with $(1,2,3,5,18,...)$ and it follows that it isn't Euclid sequence, since already $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}>2$$
