Question: 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.

Answer: 
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.
The $i$th even perfect number is $2^{n\_i-1}(2^{n\_i}-1)$, where the $n\_i$ are an enumeration of those $n$ such that $2^n-1$ is prime. In particular, the $n\_i$ are prime themselves, so that $n\_i>i$ for each $i$. Now you can see that the $i$th factor (ordered small to large) of the $i$th even perfect number is $2^{-i}$, so that the sum of the reciprocals of the $i$th factors is 2, as conjectured.