Showing that $1 - (1 - b_n)^{a_n} \sim a_n b_n$ as $n \to \infty$. Let $a_n \in \mathbb{N}_{&gt; 0}$ be an increasing sequence and $b_n \in [0,1]$ a decreasing sequence. I am interested in the behavior of the sequence $c_n \triangleq 1 - (1-b_n)^{a_n}$. Intuitively, I know that 
\begin{equation*}
1 - (1 - b_n)^{a_n} \sim a_n b_n \quad \text{as $n \to \infty$}.
\end{equation*}
However, I cannot find a rigorous proof for that. 

It is easy to upper-bound 
\begin{equation*}
1 - (1 - b_n)^{a_n} \le a_n b_n,
\end{equation*}
using the fact that $(1-x)^k \ge 1 - k x$ for $k \ge 1$ and $x$ in vicinity of $0$.

However, I am unable to find a (similar) lower-bound to conclude the claim.



I realized that I should be more clear about what I would like to conclude. In fact, I would like to prove 
\begin{equation*}
\lim_{n \to \infty} \frac1n \ln(c_n) = \lim_{n \to \infty} \frac1n \ln(a_n b_n).
\end{equation*}

As pointed out in the answer of Kim Jong Un, we can also lower-bound
\begin{equation*}
1-(1-b_n)^{a_n} \ge a_n b_n (1-b_n)^{a_n-1},
\end{equation*}
which shows,
\begin{equation*}
\frac1n \ln (c_n) \ge \frac1n \ln (a_n b_n) + \frac1n (a_n-1) \ln(1-b_n)
\end{equation*}

Now, if $\limsup_{n\to\infty} a_n b_n &lt; \infty$, 
\begin{equation*}
\lim_{n\to\infty} \frac1n (a_n-1) \ln(1-b_n) = 0
\end{equation*}
and we can conclude the claim. However, if $a_n b_n \to \infty$ the lower-bound is not useful.
Let $d=(1-b_n)\in[0,1]$, then
\begin{align*}
1-(1-b_n)^{a_n}=1-d^{a_n}&amp;=(1-d)(d^{a_n-1}+\cdots+d+1)\\
&amp;\geq(1-d)(d^{a_n-1}+\cdots+d^{a_n-1}+d^{a_n-1})\\
&amp;=(1-d)(a_nd^{a_n-1})=b_na_n(1-b_n)^{a_n-1}.
\end{align*}
Maybe logging? $(1-b_n)^{a_n} = e^{a_n \log (1-b_n)} \sim e^{-a_n b_n}$ using Maclaurin series, terms of the form $e^{O(\frac{1}{n})}$ converge to 1.