If $\varepsilon_n \rightarrow 0 $ do we have that $\sum_{n<N} \varepsilon_n b^n = o (\sum_{n<N} b^n)$

Question

If $\varepsilon_n \rightarrow 0$ and $b > 1$ do we have that $\sum_{n<N} \varepsilon_n b^n = o (\sum_{n<N} b^n)$ as $N \rightarrow +\infty$ ?

I've tried an Abel transformation but I was not able to conclude.

Answer 1

Suppose that there exists $M$ such that the terms in the sequence $\epsilon_n$ beyond the $M^\text{th}$ one all have absolute value smaller than $r$.

The finite sum $\displaystyle\sum_{n=1}^M\epsilon_n b^n$, we can regard as a constant $K$.

$|\displaystyle\sum_{n=1}^N \epsilon_n b^n|$

$\leq|K|+\displaystyle\sum_{n=M+1}^N |\epsilon_n b^n|$

$\leq|K|+\displaystyle\sum_{n=M+1}^N r b^n$

$=|K|+r b^{M+1}\displaystyle\sum_{n=0}^{N-M-1} b^n$

$=|K|+r b^{M+1}\dfrac{b^{N-M}-1}{b-1}$

The stuff in the last line is asymptotic to $ \dfrac{r(b^{N+1}-1)}{b-1}$.