Sum of a binomial type series From Binomial Theorem we know $(p+q)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!}\left( p \right)^r  \left( q \right)^{n-r}$. 
Then how to find out the sum for $\sum_{i=1}^{n} \left( 1- \sum_{r=i}^{n} \frac{n!}{r!(n-r)!}\left( p \right)^r  \left( q \right)^{n-r} \right)$ when $n$ is finite?
Furthermore, want to know the value of this summation as $n\rightarrow \infty$.
 A: The sum can be written in a nice formula, but the limit as $n$ approaches infinity is messy.
\begin{align}
\sum_{i=1}^{n} \left( 1- \sum_{r=i}^{n} \frac{n!}{r!(n-r)!} p^r  q^{n-r} \right)&=n-\sum_{i=1}^{n}  \sum_{r=i}^{n} \frac{n!}{r!(n-r)!}p ^r   q ^{n-r}
\\&=n-\sum_{r=1}^n\sum_{i=1}^r\frac{n!}{r!(n-r)!}p^r q^{n-r}
\\&=n-\sum_{r=1}^nr\frac{n!}{r!(n-r)!}p^r q^{n-r}
\\&=n-\sum_{r=1}^n\frac{n!}{(r-1)!(n-r)!}p^r q^{n-r}
\\&=n-\sum_{r=1}^n\frac{(n-1)!\,n}{(r-1)!(n-r)!}p^r q^{n-r}
\\&=n-n\sum_{r=1}^n\frac{(n-1)!}{(r-1)!(n-r)!}p^r q^{n-r}
\\&=n-n\sum_{r=0}^{n-1}\frac{(n-1)!}{r!(n-r-1)!}p^{r+1} q^{n-r-1}
\\&=n-np\sum_{r=0}^{n-1}\frac{(n-1)!}{r!\big((n-1)-r\big)!}p^r q^{(n-1)-r}
\\&=n-np(p+q)^{n-1}
\\&=n\big(1-p(p+q)^{n-1}\big).
\end{align}
$$ $$
Case 1. $p+q\gt 1.$ So $(p+q)^{n-1}\to +\infty$ as $n\to\infty.$
Subcase 1(a). $p\le 0.$  Then $n\big(1-p(p+q)^{n-1}\big)\to+\infty$ as $n\to\infty.$
Subcase 1(b). $p\gt 0.$ Then $n\big(1-p(p+q)^{n-1}\big)\to -\infty$ as $n\to\infty.$
$$ $$
Case 2. $p+q=1.$ So $(p+q)^{n-1}\to 1$ as $n\to\infty.$
Subcase 2(a). $p\lt 1.$  Then $n\big(1-p(p+q)^{n-1}\big)\to+\infty$ as $n\to\infty.$
Subcase 2(b). $p=1.$ Then $n\big(1-p(p+q)^{n-1}\big)\to 0$ as $n\to\infty.$
Subcase 2(c). $p\gt 1.$  Then $n\big(1-p(p+q)^{n-1}\big)\to-\infty$ as $n\to\infty.$
$$ $$
Case 3. $-1 \lt p+q\lt 1.$ Then $(p+q)^{n-1}\to 0$ as $n\to\infty.$  So in this case, there are no subcases, and we just get that $n\big(1-p(p+q)^{n-1}\big)\to+\infty$ as $n\to\infty.$
$$ $$
Case 4. $p+q=-1.$ So $(p+q)^{n-1}$ alternates between $+1$ and $-1$ as $n$ increases.
Subcase 4(a). $-1\lt p \lt 1.$ Then $n\big(1-p(p+q)^{n-1}\big)\to +\infty$ as $n\to\infty.$
Subcase 4(b). $p\le -1$ or $p\ge 1.$ There are various possibilities here, based on whether $p\lt -1, p=-1, -=1,$ or $p\gt 1,$ but you can check that in all those cases the terms $n\big(1-p(p+q)^{n-1}\big)$ for even $n$ behave differently for large $n$ from those for odd $n,$ so that $n\big(1-p(p+q)^{n-1}\big)$ doesn't approach a limit as $n\to\infty,$ not even $+\infty$ or $-\infty.$
$$ $$
Case 5. $p+q\lt -1.$  In this case $(p+q)^{n-1}$ alternates in sign as $n$ increases, with the values for even $n$ approaching $-\infty,$ and the values for odd $n$ approaching $+\infty.$
Subcase 5(a). $p=0.$ Then $n\big(1-p(p+q)^{n-1}\big)\to-\infty$ as $n\to\infty.$
Subcase 5(b). $p\ne 0.$ Then $n\big(1-p(p+q)^{n-1}\big)$ has no limit, with the even terms approaching either $+\infty$ or $-\infty,$ and the odd terms approaching $-\infty$ or $+\infty,$ respectively.
So we can summarize all this by writing
$$\lim_{n\to\infty}\sum_{i=1}^{n} \left( 1- \sum_{r=i}^{n} \frac{n!}{r!(n-r)!} p^r  q^{n-r}\right)=
\\\begin{cases}
\scriptsize{\text{undefined (no limit, not even }+\infty\text{ or }-\infty)},&\text{ if }(p+q\lt -1\text{ and }p\ne 0)\text{ or }(p+q=-1\text{ and either }p\le -1\text{ or }p\ge 1),
\\-\infty,&\text{ if }(p+q\gt 1\text{ and }p\gt 0)\text{ or }(p+q=1\text{ and }p \gt 1)\text{ or }(p=0\text{ and }q\lt -1),
\\0,&\text{ if }p=1\text{ and }q=0,
\\+\infty,&\text{ otherwise.}
\end{cases}$$
