How to show that the singularity of $\sum_{n=0}^{\infty} e^{nx}$ at $x=0$ is a simple pole of order 1? Consider the function
$$\sum_{n=0}^{\infty} e^{nx} = -\frac{1}{x}+\frac{1}{2}-\frac{1}{12}x \ +\dots  $$
There is an elementary "trick" to show this has as simple pole of order 1 which is equating $\sum_{n=0}^{\infty} e^{nx}$  with $\frac{1}{1-e^x}$ and then carrying out a laurent expansion of this. But i'm not really interested in this trick. Instead just I want to work explicitly with the series:
$$\sum_{n=0}^{\infty} e^{nx} = 1 + e^x + e^{2x} + e^{3x} + ... $$
And prove that the pole at $x=0$ is of order 1 directly. So a first start would be to just guess the pole has order one and then consider the expression:
$$ \lim_{x \rightarrow 0^-} \left[  x \sum_{n=0}^{\infty} e^{nx} \right] $$
Evaluating this directly leads to the expression $0 \times \infty $ which is indeterminate.
Trying to use L'Hopitals doesn't seem to be of much help either:
$$ \lim_{x \rightarrow 0} \left[  x \sum_{n=0}^{\infty} e^{nx} \right]  = \lim_{x \rightarrow 0} \frac{\sum_{n=0}^{\infty} e^{nx}}{\frac{1}{x}}$$
As you just end up differentiating the top and bottom expressions indefinitely.
Expanding as a taylor series isn't useful EITHER because the expansions come out to
$$ \sum_{n=0}^{\infty} e^{nx} = 1 + (1 + 1 + 1 ... ) + (1 + 2 + 3 + ... ) x + \frac{1}{2!} (1^2 + 2^2 + 3^2 + ... )x^2 + ... $$
Even if you knew about the Riemann Zeta function AND could account for shifting in the series you would INCORRECTLY conclude
$$  \sum_{n=0}^{\infty} e^{nx} = \frac{1}{2} - \frac{1}{12}x  + \ ... $$
And miss that singularity entirely.
So now that motivates a very natural question. Using simple first principles for real and complex analysis at the undergraduate level. How do you actually conclude that this series has a pole of order 1?
One More Idea:
We need to consider $$ \lim_{k \rightarrow \infty} \lim_{x \rightarrow 0} \left[ x \sum_{n=0}^{k} e^{nx} \right] $$
We need to find an explicit sequence $x_n \rightarrow 0$ and $k_n \rightarrow \infty$ such that $\forall \epsilon > 0, \exists \delta$ s.t.  $\forall n \in \mathbb{N}, n > \delta$
$$ \left| \left[ x_n \sum_{i=0}^{k_n} e^{ix_n} \right] - 1 \right| < \epsilon $$
 A: I guess you'd basically like a proof that $\lim_{x \rightarrow 0^-} \left[  x \sum_{n=0}^{\infty} e^{nx} \right]$ exists and is finite and nonzero, without using the geometric series formula.
This limit will look less confusing after a change of variables $r = e^x$, getting $\lim_{r \to 1^-} \ln(r) \sum_{n=0}^\infty r^n$. At this point we can once again see that it'll be easy to proceed if we used the geometric series formula, but that's not allowed! Instead, let $f(r) = \sum_{n=0}^\infty r^n$ and use l'Hopital's rule:
$$
\begin{align}
\lim_{r \to 1^-} \ln(r) f(r) &= \lim_{r \to 1^-} \frac{\ln(r)} {\frac 1 {f(r)}} \\
&= \lim_{r \to 1^-} \frac{1/r}{-f'(r) f(r)^{-2}} \\
&= -\lim_{r \to 1^-} \left(\frac{1}{r}\right) \frac{f(r)^2}{f'(r)}
\end{align}
$$
We can separately compute $$f(r)^2 = \left(\sum_{n=0}^\infty r^n\right)^2 = (1+r+r^2+r^3+\cdots)^2 = 1 + 2r + 3r^2 + \cdots = \sum_{n=1}^\infty (n-1) r^n$$
$$f'(r) = \frac{d}{dr} \left( \sum_{n=0}^\infty r^n \right) = \sum_{n=1}^\infty (n-1) r^{n-1}$$
so $f(r)^2 = f'(r)$ and thus our limit above just becomes $-\lim_{r \to 1^-} \left( \frac 1 r \right)$ which is $-1$. As desired, this is finite and nonzero, meaning that $x^1$ really was the correct power to multiply by at the start.

For simplicity, I skipped justifying a couple of steps above.
When simplifying $f(r)^2$ I "multiplied two power series together in the obvious way". This is actually valid at any point within the radius of convergence of both series (source) but it's not trivial.
Second, when simplifying $f'(r)$ I "swapped the order of $\sum_{n=0}^\infty$ and $\frac d {dr}$". Again this is true within the radius of convergence, but again it's not trivial; here's a source.
A: Here's a slightly silly answer. Note that
$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$
and so
\begin{align*}
\lim_{x \to 0^{-}} x \sum_{n=0}^\infty e^{nx}&=\lim_{x \to 0^{-}} (e^x - 1)\sum_{n=0}^\infty e^{nx} &&\text{(by the multiplication rule for limits)}\\
&=\lim_{x \to 0^-} \left[\sum_{n=1}^\infty e^{nx}-\sum_{n=0}^\infty e^{nx}\right]&&\text{(multiplying the two factors)}\\
&=\lim_{x \to 0^-} e^{0x}&&\text{(canceling like terms from each sum)}\\
&= 1
\end{align*}
(Why is this silly? Because the only reason we're not using the geometric series formula is that we're actually proving it!)
