Conceptual doubts on convergency of $a^x$.

Question

I am studying Convergency and Divergency of series in Higher Algebra by Hall and Knight. This article is particularly giving me a hard time.

To show that the expansion of $a^x$ in ascending powers of $x$ is convergent for every value of $x$.

Here $\frac{u_{n}}{u_{n-1}}=\frac{x \log _{e} a}{n-1}$; and therefore $\lim _{n=\infty} \frac{u_{n}}{u_{n-1}}<1$ whatever be the value of x; hence the series is convergent.

Here are my questions :—

Why is there $n-1$ in the denominator of $\frac{u_{n}}{u_{n-1}}=\frac{x \log _{e} a}{n-1}$.

I've copied the exact article from the book. I think the $n-1$ may be a typo.

Also,

Can anyone please explain me or at least give a rough idea of what's happening?

Any help would be appreciated.

Answer 1

Let's break the question down to smaller parts.

expansion of $a^x$ in ascending powers of $x$

presumably means (Taylor expansion):

$$a^x = 1 + (\ln a)x + \frac{\ln^2 a}{2}x^2+\dots = \sum_{n=0}^\infty \frac{((\ln a) x)^n}{n!}$$

Now, presumably

$u_n \triangleq \frac{((\ln a) x)^n}{n!}=\frac{((\ln a)x)((\ln a) x)^{n - 1}}{n(n-1)!}$

and therefore

$\frac{u_n}{u_{n-1}} = \frac{x\ln a}{n}$

which would suggest a typo indeed. Nevertheless, for the investigation of the convergence the exact form of the denominator is irrelevant. By simple change of definition ($n\geq 1$):

$u_n \triangleq \frac{((\ln a) x)^{n - 1}}{(n - 1)!} = \frac{(\ln a) x)((\ln a) x)^{n - 2}}{(n - 1)(n-2)!} $

which leads to the result as mentioned in the book.

In both cases, $\lim_{n\to\infty}\frac{u_n}{u_{n-1}} = 0.$