Binomial distribution: upper bound and lower bound This question is discussed in  Feller's Introduction to Probability Theory and Its Applications and i am relatively new in probability theory, so this theme is incomprehensible for me:
The probability at least $r$ successes is:
$$P(S_n \ge r)=\sum_{v=0}^\infty b(r+v;n,p).$$
The formula for upper bound is:
$$P(S_n \ge r)≤\frac{rq}{(r-np)^2} : \quad r≥np $$ 
the formula for lower bound is:
$$P(S_n \le r)≤ \frac{(n-r)p}{(np - r)^2} : \quad r≤np$$
But i don't understand how the author derived such formulas.
So, if it's possible, can you please explain the meaning and derivation of formulas in a simple(not complicated) way?
Thank you
 A: To fill in some details, we have (from Feller, pg. 151):
\begin{eqnarray*}
&& S_n \sim B(n,p) \\
&& q = 1-p \\
&& b(k;n,p) := P(S_n=k).
\end{eqnarray*}
I'll try to flesh out Feller's argument. Firstly, for $r\gt np,\;$ we look at ratios of probabilities:
\begin{eqnarray*}
\dfrac{P(S_n=r+1)}{P(S_n=r)} &=& \dfrac{\binom{n}{r+1}p^{r+1}q^{n-r-1}}{\binom{n}{r}p^{r}q^{n-r}} \\
&& \\
&=& \dfrac{\dfrac{n!}{(r+1)!(n-r-1)!}p^{r+1}q^{n-r-1}}{\dfrac{n!}{r!(n-r)!}p^{r}q^{n-r}} \\
&& \\
&=& \dfrac{(n-r)p}{(r+1)q} \\
&& \\
&\leq& \dfrac{(n-r)p}{rq} \\
&& \\
&=& 1 - \dfrac{rp-np+rq}{rq} \\
&& \\
&=& 1 - \dfrac{r-np}{rq}\qquad\qquad\qquad\qquad\qquad\text{(1)} \\
&\lt& 1\qquad\text{since } r\gt np.
\end{eqnarray*}
So if $k\gt r$, then
\begin{eqnarray*}
P(S_n=k) &=& \dfrac{P(S_n=k)}{P(S_n=k-1)} \dfrac{P(S_n=k-1)}{P(S_n=k-2)} \cdots \dfrac{P(S_n=r+1)}{P(S_n=r)}P(S_n=r) \\
&& \\
&\leq& P(S_n=r)\left(1 - \dfrac{r-np}{rq} \right)^{k-r} \qquad\qquad\qquad\text{using (1)}.
&& \\
\therefore\quad P(S_n\geq r) &=& \sum_{k=r}^{\infty} P(S_n=k) \\
&& \\
&\leq& \sum_{k=r}^{\infty} P(S_n=r) \left(1 - \dfrac{r-np}{rq} \right)^{k-r} \\
&& \\
&=& P(S_n=r) \sum_{k=0}^{\infty} \left(1 - \dfrac{r-np}{rq} \right)^{k} \\
&& \\
&=& P(S_n=r) \dfrac{1}{1 - \left(1 - \dfrac{r-np}{rq} \right)} \qquad\text{using $\sum_{m=0}^\infty x^m=\dfrac{1}{1-x}$ if $0\lt m\lt 1$} \\
&& \\
&=& P(S_n=r)\dfrac{rq}{r-np}.\qquad\qquad\qquad\qquad\qquad\text{(2)}
\end{eqnarray*}
We now find an upper bound for $P(S_n=r)$. From $(1)$ we know that if $np\leq k\leq r$, then $P(S_n=k) \geq P(S_n=r)$.
So there are at least $r-np$ such numbers. (Note: $np$ may not be an integer.) Therefore,
$$1\geq \sum_{np\leq k\leq r} P(S_n=k) \geq \sum_{np\leq k\leq r} P(S_n=r) \geq (r-np)P(S_n=r).$$
Therefore, $P(S_n=r) \leq \dfrac{1}{r-np}$. Substituting this into $(2)$, we arrive at
$$P(S_n \geq r) \leq \dfrac{rq}{(r-np)^2}.$$
The second result derives directly from this by mapping:
\begin{eqnarray*}
p &\mapsto& q \\
q &\mapsto& p \\
r &\mapsto& n-r.
\end{eqnarray*}
