Convergence of partial sums and their inverses 
If a sequence $s_{k}$ of partial sums converges to a nonzero limit, and we assume that $s_{k} \neq 0$  for all $k$  $\epsilon$ $\mathbb{N}$, then also the sequence $\left \{ \frac{1}{s_{k}}  \right \}$ converges. 

In my book, $s_{k}$ is defined as $\sum_{j = 1}^{k}\frac{a_{j}}{10^{j}}$
which is a decimal expansion. 
I can't immediately see why this sequence converges - Maybe I'm just braindead today, but I can't think of any examples in my head that are making sense to me. Can anyone point me in the right direction about why this is true? 

In general, if $\{a_n\}$ is any convergent sequence with a limit $a\neq 0$, then $\dfrac{1}{a_n}$ converges to $\dfrac{1}{a}$.
Proof. Let $\epsilon>0$. Then 
$$
\left|\frac{1}{a_n}-\frac{1}{a}\right|=\left|\frac{a-a_{n}}{aa_n}\right|
$$
Since $a_n\to a$ as $n\to\infty$, we can choose a positive integer $N_1$ such that if $n\ge N_1$, then $|a-a_n|\leq \dfrac{|a|}{2}$ for all $n\ge N_1$. This implies $a_n\ge\dfrac{|a|}{2}$ for all $n\ge N_1$. Now choose a positive integer $N_2$ such that $|a-a_n|\leq \dfrac{a^2}{2}\epsilon$ for all $n\ge N_2$. Let $N=\max(N_1, N_2)$. It follows that 
$$
\left|\frac{1}{a_n}-\frac{1}{a}\right|=\left|\frac{a-a_{n}}{aa_n}\right|=\frac{|a-a_n|}{|a||a_n|}\leq \frac{\frac{a^2}{2} \epsilon}{\frac{a^2}{2}}=\epsilon
$$
for each $n\ge N$. So $\dfrac{1}{a_n}\to\dfrac{1}{a}$ as desired.
