In chapter VI, Introduction to Probability theory and its applications, Feller obtains an upper bound for the the right tail and left tail of the binomial distribution.
Theorem. If $r \geq np$, the probability of atleast $r$ successes is bounded above by
\begin{align*} P(S_n \geq r) \leq b(r;n,p) \frac{(r+1)q}{(r+1)-(n+1)p} \tag{1} \end{align*}
He uses this upper bound in the proof of the fact that if trials are stochastically independent, and if $S_n$ is the number of successes in $n$ trials, then the average number of successes, $\frac{S_n}{n}$ approaches $p$.
We are interested to find the chance that the number of successes $S_n$ exceeds $n(p+\epsilon)$, where $\epsilon > 0$ is an arbitrary positive real number. Since, $r \geq n(p + \epsilon) > np$, we can use the result derived in (1).
He then finds a stable upper bound as below.
So, I am not able to obtain the same upper bound as (4.1). But, I would like to ask, if my below work checks out and sufficiently rigorous.
Proof (My attempt).
Since $r \leq n$ and $r \geq n(p+\epsilon)$, we have:
\begin{align*} \frac{(r+1)q}{(r+1)-(n+1)p} &\leq \frac{(n+1)q}{(n(p+\epsilon)+1)-(n+1)p}\\ &=\frac{(n+1)q}{np + n\epsilon + 1 - np - p} \\ &=\frac{(n+1)q}{n\epsilon + q} \end{align*}
So, we proved that the $n$th term of the sequence $(p_n)$ has an upper bound:
\begin{align*} P\{S_n > n(p + \epsilon)\} \leq b(r;n,p) \cdot \frac{(n+1)q}{n\epsilon + q} \end{align*}
Since probabilities are non-negative,
\begin{align*} 0 \leq P\{S_n > n(p + \epsilon)\} \leq b(r;n,p) \cdot \frac{(n+1)q}{n\epsilon + q} \end{align*}
Passing to the limit, as $n \to \infty$, we have:
\begin{align*} \lim 0 \leq \lim P\{S_n > n(p + \epsilon)\} \leq \lim \left[ b(r;n,p) \cdot \frac{(n+1)q}{n\epsilon + q}\right] \end{align*}
Now,
\begin{align*} \lim \left[b(r;n,p) \cdot \frac{(n+1)q}{n\epsilon + q}\right] &= \lim b(r;n,p) \cdot \lim \frac{\left(1+\frac{1}{n}\right)q}{\epsilon + \frac{q}{n}}\\ &= \lim b(r;n,p) \cdot \frac{q}{\epsilon} \end{align*}
It's easy to see, that as $n \to \infty$, $b(r;n,p) \to 0$. Note that, as $n \to \infty$, $r \to \infty$, because $r \geq n(p + \epsilon)$.
\begin{align*} \lim b(r;n,p) &= \lim \frac{n!}{(n-r)!r!} \cdot \lim p^r q^{n-r}\\ &\leq \lim \frac{n!}{n!0!} \cdot \lim p^n q^n \quad \{ r \ge 0 \text{ and } r \leq n \} \\ &= \lim p^n q^n \{ \text{ since } (q^n) \to 0, \text{ if } |q|<1 \}\\ &= 0 \end{align*}
Since $b(r;n,p) \geq 0$, $\lim b(r;n,p) = 0$.
By the squeeze theorem,
\begin{align*} \lim P(S_n \geq n(p + \epsilon)) = 0 \end{align*}