## Definition 3.5.2: Exponent Sequence

**Exponent Sequence**: The convergence depends on the size of the exponent

*a:*

*a > 0*: the sequence diverges to positive infinity*a = 0*: the sequence is constant*a < 0*: the sequence converges to 0

**Exponent sequence with**

*a = 2*

**Exponent sequence with a = -2**

### Proof:

Write*n*. Then:

^{a}= e^{a ln(n)}- if
*a > 0*then as*n*approaches infinity, the function*e*approaches infinity as well^{a ln(n)} - if
*a < 0*then as*n*approaches infinity, the function*e*approaches zero^{a ln(n)} - if
*a = 0*then the sequence is the constant sequence, and hence convergent

But actually we first need to understand exactly what
*n ^{a}* really means:

- If
*a*is an integer, then clearly*n*means to multiply^{a}*n*with itself*a*times - If
*a = p/q*is a rational number, then*n*means to multiply^{p/q}*n**p*times, then take the*q*-root - It is unclear, however, what
*n*means if^{a}*a*is an irrational number. For example, what is*n*, or^{}*n*?^{}

*n*for all

^{a}*a*is to resort to the exponential function:

In that light, the original 'proof' was not bad after all, but of course we now need to know how exactly the exponential function is defined, and what its properties are before we can continue. As it turns out, the exponential function is not easy to define properly. Generally one can either define it as the inverse function of then^{a}= e^{a ln(n)}

*ln*function, or via a power series.

Another way to define *n ^{a}* for

*a*being rational it to take a sequence of rational numbers

*r*converging to

_{n}*a*and to define

*n*as the limit of the sequence

^{a}*{n*. There the problem is to show that this is well-defined, i.e. if there are two sequences of rational numbers, the resulting limit will be the same.

^{rn}}In either case, we will base our proof on the simple fact:

ifwhich seems clear enough and could be formally proved as soon as the exponential function is introduced to properly definep > 0andx > ythenx^{p}> y^{p}

*x*.

^{p}
Now take any positive number *K* and let *n* be
an integer bigger than *K ^{1/a}*. Since

we can raise both sides to then > K^{1/a}

*a*-th power to get

which means - sincen^{a}> K

*K*was arbitrary - that the sequence

*{n*is unbounded.

^{a}}
The case *a = 0* is clear (since *n ^{0} = 1*).
The second case of

*a < 0*is related to the first by taking reciprocals (details are left as an exercise). Since we have already proved the first case we are done.