<s>[INST] Convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$. 
Show that $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$$ converges for every $x>1$.

let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
 [/INST] Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$\begin{array}\\
s_m(y)
&=\sum_{n=1}^{m} (-1)^{n-1}(-y)^n\\
&=\sum_{n=1}^{m} (-1)^{n-1}(-1)^ny^n\\
&=-\sum_{n=1}^{m} y^n\\
&=-y\sum_{n=0}^{m-1} y^n\\
&=-y\dfrac{1-y^m}{1-y}\\
&=\dfrac{-y}{1-y}-\dfrac{-y^{m+1}}{1-y}\\
\end{array}
$
so
$\begin{array}\\
s_m(y)+\dfrac{y}{1-y}
&=\dfrac{y^{m+1}}{1-y}\\
\end{array}
$
Therefore
$\sum_{n=1}^{m} \frac{(-1)^{n-1}}{x^n}+\dfrac{-1/x}{1+1/x}
=\dfrac{1}{(-x)^{m+1}(1+1/x)}
$
or
$\sum_{n=1}^{m} \frac{(-1)^{n-1}}{x^n}-\dfrac{1}{x+1}
=\dfrac{(-1)^{m+1}}{x^{m}(x+1)}
$.
What is needed now is
to show that
$\lim_{m \to \infty} \dfrac{1}{x^{m}(x+1)}
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
\ge 1+mz
\gt mz
=m(x-1)$,
so
$ \dfrac{1}{x^{m}(x+1)}
\lt \dfrac{1}{m(x-1)(x+1)}
=\dfrac{1}{m(x^2-1)}
$,
so to make
$\dfrac{1}{x^{m}(x+1)}
\lt \epsilon
$
it is enough to take
$m
\gt \dfrac1{\epsilon(x^2-1)}
$.
This is certainly
not the best $m$,
but it is
completely elementary.
</s>