I am studying and solving through the problems from the excellent monograph by Feller - Introduction to probability theory and applications. I am having some trouble understanding the derivation of the chapter VI.4 - the law of large numbers.
On page 159, the author derives an expression for an upper bound for the ratio of two successive binomial terms $b(k;n,p)/b(k-1;n,p)$. For $k \ge (r+1)$:
\begin{align*} \frac{(n+1)p - pk}{kq} &\le \frac{(n-r)p}{(r+1)q}\\ \frac{b(k;n,p)}{b(k-1;n,p)}&\le \frac{(n-r)p}{(r+1)q} \end{align*}
Set herein, $k=r+1, \ldots, r+\nu$ and multiply the $\nu$ inequalities to obtain: \begin{align*} \frac{b(r+1;n,p)}{b(r;n,p)} &\le \frac{(n-r)p}{(r+1)q}\\ \frac{b(r+2;n,p)}{b(r+1;n,p)} &\le \frac{(n-r)p}{(r+1)q}\\ \implies\frac{b(r+2;n,p)}{b(r+1;n,p)}\cdot \frac{b(r+1;n,p)}{b(r;n,p)} &\le \left(\frac{(n-r)p}{(r+1)q}\right)^2\\ \implies\frac{b(r+2;n,p)}{b(r;n,p)}&\le \left(\frac{(n-r)p}{(r+1)q}\right)^2\\ \implies\frac{b(r+\nu;n,p)}{b(r;n,p)}&\le \left(\frac{(n-r)p}{(r+1)q}\right)^\nu \end{align*}
If we let this ratio, be less than unity $1$, that is : \begin{align*} \frac{(n-r)p}{(r+1)q}& < 1\\ \implies (n-r)p &< (r+1)q\\ \implies np - rp &< (r+1)(1-p)\\ \implies np - rp &< (r+1) - (r+1)p\\ \implies np - rp &< (r+1) - rp - p\\ \implies np &< (r+1) - p\\ \implies (n+1)p &< (r+1)\\ \implies (r+1) &> np + p \end{align*}
Question. How do I conclude that $r \ge np$?
For $r \ge np$, the fraction in the braces is less than unity, and the summation over $\nu$ leads to a finite geometric series with the ratio $(n-r)p/(r+1)q$. We conclude that for $r \ge np$, it must be bounded above by the infinite series: \begin{align*} \frac{\sum_{\nu=0}^{n-r}b(r+\nu;n,p)}{b(r;n,p)} &\le 1+ \left(\frac{(n-r)p}{(r+1)q}\right) \\ &+ \left(\frac{(n-r)p}{(r+1)q}\right)^2 + \ldots \end{align*}
We can therefore write :
\begin{align*} \sum_{\nu=0}^{n-r}b(r+\nu;n,p) &\le b(r;n,p)\cdot \frac{1}{1-\frac{(n-r)p}{(r+1)q}}\\ &=b(r;n,p)\cdot \frac{(r+1)q}{(r+1)q-(n-r)p} \tag{1} \end{align*}
Theorem. If $r \ge np$, the probability of atleast $r$ successes satisfies the above inequality.
If $S_\nu$ is the number of successes in $n$ identical Bernoulli trials, $S_n/n$ is the average number of successes. Our intuitive notion of probability says, that $S_n/n \to p$, where $p$ is the probability of successes.
Consider, for example, the probability that $S_n/n$ exceeds $p + \epsilon$, where $\epsilon > 0$ is arbitrarily small but fixed. This probability is the same as $P(S_n > n(p + \epsilon)$ which equals the left side of (1). Then $(1)$ implies:
\begin{align*} P\{S_n > n(p + \epsilon)\} < b(r;n,p) \frac{n(p+\epsilon)+q}{n{\epsilon} + q} \end{align*}
I am unable to derive the expression on the right hand side. How does that come about?