limit of double binomial sum Prove that
$$\lim_{\max(M,N) \to \infty} \frac{\sum_{i=0}^{M-1} \sum_{j=0}^{N-1} p^{i} (1-p)^{j} {i+j \choose i}}{\min (\frac{M}{p}, \frac{N}{1-p})} = 1 $$
where $0<p<1$ and $M, N$ are positive integers.
I did it for the case that one of $M, N$ is fixed by using the differentiation of power series representation of $\frac{1}{1-p}$, but I have no idea about the remaining case - both of $M, N$ go to infinity.

My work - Proof for the special case where $N$ is finite:
The original problem becomes $$\lim_{M \to \infty} \sum_{i=0}^{M-1} \sum_{j=0}^{N-1} p^{i} (1-p)^{j} {i+j \choose i} = \frac{N}{1-p}. $$
To see this, for each $j$,
$$(1-p)^{j} \sum_{i=0}^{\infty} p^{i}  {i+j \choose i} = (1-p)^{j} \frac{1}{j!} \frac{d^{j}}{dp^{j}} \big( \frac{1}{1-p} \big) = \frac{(1-p)^{j}}{(1-p)^{j+1}} = \frac{1}{1-p}, $$
whence
$$\lim_{M \to \infty} \sum_{i=0}^{M-1} \sum_{j=0}^{N-1} p^{i} (1-p)^{j} {i+j \choose i}  = N \cdot \frac{1}{1-p} = \frac{N}{1-p}. $$
 A: Since this looks like a combinatorial problem (with all those sums and binomials), it makes sense to try to find a combinatorial interpretation of the values. In particular, define: $$f(M,N)=\sum_{i=0}^{M-1}\sum_{j=0}^{N-1}p^i(1-p)^j{i+j\choose i}$$
Before we get into what this means, we notice a useful quality of this function:
$$f(M,N)=1+pf(M-1,N)+(1-p)f(M,N-1)$$
which can be proven since it expands to:
$$f(M,N)=1+\left(\sum_{i=0}^{M-2}\sum_{j=0}^{N-1}p^{i+1}(1-p)^{j}{i+j\choose i}\right)+\left(\sum_{i=0}^{M-1}\sum_{j=0}^{N-2}p^{i}(1-p)^{j+1}{i+j\choose i}\right)$$
which, after we jigger with the sum limits becomes
$$f(M,N)=1+\left(\sum_{i=1}^{M-1}\sum_{j=0}^{N-1}p^{i}(1-p)^{j}{i+j-1\choose i-1}\right)+\left(\sum_{i=0}^{M-1}\sum_{j=1}^{N-1}p^{i}(1-p)^{j}{i+j-1\choose i}\right)$$
and combining the sums where $M,N\geq 1$, splitting off the sums where $i=0$ or $j=0$ separately, and using the identity that ${i+j-1\choose i-1}+{i+j-1\choose i}={i+j\choose i}$ gives
$$f(M,N)=1+\left(\sum_{i=1}^{M-1}p^{i}(1-p)^{0}{i+0-1\choose i-1}\right)+\left(\sum_{j=1}^{N-1}p^{0}(1-p)^{j}{0+j-1\choose 0}\right)+\left(\sum_{i=1}^{M-1}\sum_{j=1}^{N-1}p^{i}(1-p)^{j}{i+j\choose i}\right)
$$
but since ${i+0-1\choose i-1}=1={i+0\choose i}$ and ${0+j-1\choose 0}=1={0+j\choose 0}$ and $1=p^0(1-p)^0{0\choose 0}$, the above all collapses to
$$f(M,N)=\sum_{i=0}^{M-1}\sum_{j=0}^{N-1}p^i(1-p)^j{i+j\choose i}$$
as we expected. This all goes to ensure that we have the identity
$$f(M,N)=1+pf(M-1,N)+(1-p)f(M,N-1).$$
Why is this important? Well, I interpret $f$ as follows:

If a person starts at the position $(M,N)$ and chooses to walk left (decreasing the $x$-coordinate by $1$) with probability $p$ or down (decreasing the $y$-coordinate) with probabiliy $(1-p)$, then $f(M,N)$ is the expected number of steps before one of the coordinates is $0$.

This is obvious from the recursive definition - the sum can be read that the person will take $1$ step (hence the $1+\ldots$ term) and then will, with probability $p$ be in the state $(M-1,N)$ and with probability $(1-p)$ be in the state $(M,N-1)$.
An intuitive way to see all of the above is that $p^i(1-p)^j{i+j\choose j}$ is the probability that the person would walk through $(0,0)$ if they started at $(i,j)$ that - or equivalently, it is the expected number of times that they would walk through $(M-i,N-j)$ if they started at $(M,N)$. If we sum up all of these expected values, we get the expected number of steps the person would would spend walking before reaching the $x$ or $y$ axes.
To further elucidate the problem, define two random variables $T_x$ and $T_y$, representing the number of steps taken before the $x$ and $y$ coordinates respectively are $0$, given that the person starts at $(M,N)$. Then, we can write three suggestive relations with these (where $\mathbb{E}$ is the expectation operator):
$$\mathbb{E}(T_x)=\frac{M}p$$
$$\mathbb{E}(T_y)=\frac{N}{1-p}$$
$$\mathbb{E}(\min(T_x,T_y))=f(M,N)$$
where the last one is because the condition of the person having one zero coordinate occurs whenever either $T_x$ or $T_y$ does. We're essentially being asked to prove that, in this case
$$\mathbb{E}(\min(T_x,T_y))\text{ is asymptotic to }\min(\mathbb{E}(T_x),\mathbb{E}(T_y)).$$
What is obvious from this is that, since the minimum of two variables must have expected value less than either variable, we get
$$f(M,N)\leq \min\left(\frac{M}p,\frac{N}{1-p}\right).$$
If we can show that, for any $\varepsilon$ and all large enough $M$ and $N$, that
$$(1-\varepsilon)\min\left(\frac{M}p,\frac{N}{1-p}\right)\leq f(M,N)$$
then we have shown that the limit converges to $1$, as expected.
To do this, consider the probability, for fixed $\lambda_1$ of
$$P\left(T_x \leq (1-\lambda_1)\frac{M}p\right)$$
that is that the $x$ coordinate reaches $0$ by a constant factor faster than expected. We can easily show that this is the same as saying probability as the probability that, after $(1-\lambda_1)\frac{M}p$ steps, the $x$ coordinate of the person is $0$ or less - now, the $x$-coordinate is distributed according to a translation of binomial distribution with mean $\lambda_1 M$ and the probability that this is less than $0$ will clearly shrink to $0$ as $M$ increases to infinity (as a consequence of the central limit theorem). Thus, for any $\lambda_2>0$, we can choose $M$ large enough so that $P\left(T_x \leq (1-\lambda_1)\frac{M}p\right)<\lambda_2$. By a similar argument, we can choose $N$ large enough so that $P\left(T_y \leq (1-\lambda_1)\frac{N}(1-p)\right)<\lambda_2$.
What do we do with this knowledge? Well, this means that
$$p=P\left(\min(T_x,T_y)\leq (1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)\right)<2\lambda_2$$
since at least one of the before conditions must hold for this to hold. However, note that since $\min(T_x,T_y)>0$ we know that the function
$$r(x)=\begin{cases}0&&\text{if }x\leq (1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)\\(1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)&&\text{if }x>(1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)\end{cases}$$
will round every value of $\min(T_x,T_y)$ downwards implying:
$$f(M,N)=\mathbb{E}(\min(T_x,T_y))\geq\mathbb{E}(r(\min(T_x,T_y)))$$
but since $r$ would be $0$ with probability $p$ and would be $(1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)$ with probability $(1-p)$, this yields that
$$\begin{align*}\mathbb{E}(r(\min(T_x,T_y)))&=(1-p)(1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)\\&>(1-2\lambda_2)(1-\lambda_1)\min\left(\frac{M}p,\frac{N}{1-p}\right)\end{align*}$$
Choose $\lambda_1$ and $\lambda_2$ such that $(1-2\lambda_2)(1-\lambda_1)>(1-\varepsilon)$ and we have already demonstrated that, for sufficiently large $M$ and $N$, the value $f(M,N)$ is greater than the above. This means that your limit does, indeed, converge to $1$.
