Exponential Series

[This is a mathy post. Beware.]

I'm sure someone has already reasoned through this somewhere before, but whatever.

Consider the sequence defined by the following exponential function:

where a is a real number. (Of course, in its general form, there could be a constant term in front of a which is unaffected by the exponent. However, the constant term is irrelevant in the forthcoming analysis, so it is ignored for simplicity.) We want to examine the relationship between the n^th term of the sequence and the series formed by the first n-1 terms of the sequence. Specifically, is the following statement true?


Put into English, this inequality asserts the following: the n^th term of an exponential sequence is greater than the sum of all the previous terms in the sequence.

* * *

Proposition:


Proof: by induction on n

Base case: n = 1
The inequality simplifies to a > 1, which is true because a is greater than or equal to 2.

Inductive step:
Assume the hypothesis is true for all integers 1,...,n. Show that it holds for n + 1; that is,

This can alternatively be written as:


The left hand side (LHS) of the inequality is at least double aⁿ, since a is greater than or equal to 2. The right hand side (RHS) of the inequality is strictly less than double aⁿ, since the summation term is strictly less than aⁿ by the inductive hypothesis. Therefore, the LHS of the inequality is strictly greater than the RHS, and the proposition is proved.

* * *

It was essential that our base a be at least 2 for the inequality to be true. To demonstrate why this is so, consider the case where a = 1. The LHS of the inequality will always be the value 1 for any value of n. The RHS will always be n, since it is the summation of n exponential terms with base 1. Thus, the inequality does not hold in this case.

So the inequality is always true when a = 2, and never true when a = 1. However, how does the inequality hold up when 1 < a < 2? The answer is...it depends! Both on what the value of the base a is, as well as what the value of n is!

As an example, suppose a = 1.9.
When n = 1, the LHS is 1.9 and the RHS is 1, and the inequality is true.
When n = 2, the LHS is 3.61 and the RHS is 2.9, and the inequality is true.
When n = 3, the LHS is 6.859 and the RHS is 6.51, and the inequality is true.
When n = 4, the LHS is 13.0321 and the RHS is 13.369, and the inequality is false.

It appears that when 1 < a < 2, the inequality "flips" at a certain value of n. We will call this value of n the "critical n". In fact, the border cases of a = 1 and a = 2 each have a critical n. When a = 1, the critical n is 1, which implies that the inequality flips immediately and is never true; and when a = 2, the critical n is infinity, which implies that the inequality never flips and is always true in its originally presented form.

We want to analytically derive a formula that will reveal for what value of n the inequality no longer holds when 1 < a < 2. We can do this by using the formula for the sum of a geometric series with a finite number of terms:


Substituting this sum for the RHS of the inequality, we are attempting to find the largest integer value of n such that:


Here's a chance to brush up on your algebra skills...


Note that the inequality flips in Step 2. This occurs because the denominator in Step 1 is negative (recall that 1 < a < 2).

The critical n is the smallest integer such that the inequality in Step 7 is no longer true (or one more than the largest integer such that the inequality is still true). So the critical n can be determined by computing the RHS of the inequality in Step 7, and then taking the ceiling of that value (that is, rounding up to the nearest integer).

We see that this formula indeed holds up in the example given above when a = 1.9. The RHS of the inequality in Step 7 is 3.587, which is 4 when rounded up. Indeed, it also holds up when a = 1 and when a = 2, as the numerator is 0 and infinity in each of those cases, respectively. These round up to 1 and infinity, respectively. (Technically the RHS of the inequality in Step 7 is in the indeterminate form 0/0 when a = 1, but for the sake of the argument, we'll ignore this and only consider the numerator value.)

It is also possible to determine the smallest value of the base a for which a given critical n applies (i.e., a fixed n). Unfortunately, there is not a closed-form solution for a as for the critical n; rather, a must be found as a solution to an equation. However, some simple rearrangement reveals that this equation takes on a relatively simple polynomial structure:


We can plot the critical n as a function of a between 1 and 2. It is naturally a step function since n only takes integer values.



As expected, as a approaches 2, the critical n approaches infinity.

* * *

Conclusion: We have found the conditions under which the n^th term of an exponential sequence is greater than the sum of all the previous terms in the sequence. If the base a is 2 or greater, then this is always true. If the base a is 1 or lower, then this is never true. But for bases a between 1 and 2, this is only true for values of n up until a given "critical n". We have found a formula that can determine what this "critical n" is for a given base, as well as an implicit formula that can determine the base a at which a given n is the critical n.