Following [this](https://math.stackexchange.com/questions/1566087/can-the-existence-of-infinitely-many-even-perfect-numbers-be-settled-by-a-diagon/1566168?noredirect=1#comment3188832_1566168), let's define the notion of perfect sequence as follows:


$(u\_{i})\_{i}$ is a perfect sequence if and only if it is the sequence of divisors of an even perfect number in increasing order or the infinite sequence $(u\_{i})\_{i}, \ \ u\_{i}=2^{-i}$.


Let's now list the set of finite perfect sequences by increasing number of terms, so that $(v\_{1,i})=(1, 2, 3, 6)$, $(v\_{2,i})=(1, 2, 4, 7, 14, 28)$, And so on.


My question is: is the diagonal sequence $(v\_{i,i})\_{i}$ such that $\sum\_{i}{v\_{i,i}}^{-1}=2$?


Many thanks in advance.



Diagonal argument for even perfect numbers
You are asking if a sum of $i$ numbers equals $2$.  

The partial sum  

$S\_k=\frac{1}{u\_{1,1}}+...+\frac{1}{u\_{k,k}}$ does not change as $i$ grows for every $k<i$ so, if you want to reach $2$ there are these possibilities :  

**1.** $i=\infty$  

**2.** $i<\infty$  

We don't know if there are infinitely or finitely many even perfect numbers so we cannot expect to answer your question.  

It is like you are asking if the sum of the reciprocals of even perfect numbers converges to a rational.  

If it does, there are finitely many,but we do not have any idea at all about how many exist.  

But in your comment above, it seems that you understand what I am saying. 


