Consider a positive sequence $x\_n >0$ that satisfy the condition that there exists a constant $0<\alpha<1$ such that $x\_{n+1} \geq \alpha (x\_1+\ldots{} +x\_{n})$.


What can be said about the set of all such sequences? Can one solve for them in terms of a formula for $x\_n$?


I am particularly interested in knowing whether or not all sequences that grow at least exponentially fast in or faster must necessarily satisfy the inequality stated in the question.



Sequences satisfying $x_{n+1} \geq \alpha (x_1+\ldots{} +x_{n})$ for $\alpha<1$
Not very much can be said about all such sequences. In essence all that can be said is that they grow at least exponentially. As previous comments have noted, for all $n$, we have $x\_n \geq y\_n$, where $y\_1 := x\_1$ and for all $n>1$,
$$
y\_n := \alpha \sum\_{i=1} ^{n-1} y\_{i} = \alpha \sum\_{i=1} ^{n-2} y\_{i} + \alpha y\_{n-1} = (1+\alpha) y\_{n-1}.
$$
Therefore, $y\_n = \alpha (1+\alpha)^{n-2} x\_1$, which is valid for all $n\geq 2$. Thus if $x\_n$ satisfies the desired property in the original post, then for all $n\geq 2$, we have $x\_n \geq \alpha (1+\alpha)^{n-2} x\_1$, and this lower bound is of course obtained by the sequence $y\_n$.


To strengthen this bound somewhat, for all $n > m \geq 1$, we must have
$$
x\_{n} \geq \alpha (1+\alpha)^{n-m-1}\left(\sum\_{i=1} ^{m} x\_{i} \right) \geq x\_m \alpha (1+\alpha)^{n-m-1}.
$$
Therefore, if $x\_m \neq 0$ and $t\geq 1$, we have
$$
\frac{x\_{m+t}}{x\_m} \geq \alpha (1+\alpha)^{t-1},
$$
which implies that for all sufficiently large $t$ (which depends only on $\alpha$), $x\_{m+t} > x\_m$ and that $x\_n$ goes to infinity.




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However, trying to say something much more precise is difficult. Sequences that have the desired property you described certainly grow fast ($x\_n \geq \alpha x\_1 (1+\alpha)^{n-2}$), but it is not the case that all sequences that grow at least this fast must have your desired property. Consider for example the sequence $a\_n = (2^1, 4^2, 2^3, 4^4, 2^5, 4^6, ...).$ Then $a\_n \geq a\_1 2^{n-1}$, but there is no alpha for which your desired property holds.


This problem arose because some of the $a\_i$ grew more slowly than the others (but honestly, this difference in growth rate wasn't even that bad)! If you know that $\gamma \_n$ satisfies
$$
c\_1 R^n \leq \gamma\_n \leq c\_2 R^n,
$$
then you can in fact conclude that $\gamma\_n$ satisfies a bound such as you desired.




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Hoping to find a closed formula is of course as hopeless as ever even assuming a sequence has your desired property. This is because you can start with any sequence $A(n)$ that is perhaps as complicated and difficult as possible. Then if $B(n)$ is a function that grows sufficiently quickly (e.g., iterating $n^{n}$ a few times), then $A(n) + B(n)$ will have your desired property, but finding a closed formula for it will be at least as hard as finding a closed formula for $A(n)$ [which could be arbitrarily difficult].




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In short, knowing more about your particular sequence may help a lot, but in general, your condition says very very little about the sequence (other than the fact that it grows fast).


