Can one have $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n10^{k-n}(-a)^{k-1}= 0$ for some $\lvert a \rvert > 1/10$?

Question

I've used the geometric series formula to show that $$ \lim_{n\to\infty}\sum_{k=1}^{n} 10^{k-n}(-a)^{k-1}=\lim_{n\to\infty}\Big[10^{1-n}\sum_{k=1}^{n} (-10a)^{k-1}\Big]=\Big(\lim_{n\to\infty}10^{1-n}\Big)\frac{1}{1+10a} $$ for $\mid{a}\mid<\frac{1}{10}$ which evaluates to zero due to the $10^{1-n}$ term, but I was wondering if there's a way to prove this for some larger $\mid{a}\mid$?

Answer 1

\begin{align} \sum_{k=1}^{n} 10^{k-n}(-a)^{k-1}&=\Big[10^{1-n}\sum_{k=1}^{n} (-10a)^{k-1}\Big] \\&= \Big[10^{1-n} \frac{(1-(-10a)^n)}{1-(-10a)}\Big] \\&=\frac{10}{1+10a} \left(\frac{1}{10^n}- (-a)^n \right) \end{align}

In particular, I can choose $a= 0.5$ and we can see that as $n \to \infty$, the sum converges to $0$.