The limit of binomial distributed random variable Edit
(As Robert pointed out, what I was trying to prove is incorrect.  So now I ask the right question here, to avoid duplicate question)
For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial.  Intuitively, we know if $p > 0$,  $\Pr \{N < \infty \} = 0$, in other word $N \rightarrow \infty$.  But I got stuck when I try to prove it mathematically.
\begin{aligned}
\Pr \{ N < \infty \} 
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
\end{aligned}
I've totally no idea how to calculate the last expression.

(Original Question)
For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial.  Can we prove that $\Pr \{N < \infty \} = 1$  by:
\begin{aligned}
\Pr \{ N < \infty \} 
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
& = \lim_{m \rightarrow \infty}\sum_{i=1}^{m} b(i; m, p) \\
& = \lim_{m \rightarrow \infty}[p + (1 - p)]^m \\
& = \lim_{m \rightarrow \infty} 1^m \\
& = 1
\end{aligned}
I know there must be some mistake in the process because if $p = 1$, N must infinite. So the  equation only holds when $ p < 1 $.  Which step is wrong?

As long as $p > 0$, $N$ will be $\infty$ with probability 1.  The first mistake is in
$$ \sum_{i=1}^\infty b(i;\infty,p) = \lim_{m \to \infty} \sum_{i=1}^m b(i; m,p)$$
