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Consider the following recurrence relation in two variables: $$f(a, b) = \frac{a}{a+b} f(a-1,b)+ \frac{b}{a+b}f(a+1,b-1) $$ for positive integers $a$ and $b$, with the boundary conditions $f(0,b)=0$ for integers $b>0$ and $f(a,0)=1$ for integers $a\ge0$.

I believe that $f(1,n)$ converges to $1$ as $n$ tends to infinity. However I can't see how to prove it.

Numerically the convergence is slow.

Consider the following recurrence relation in two variables: $$f(a, b) = \frac{a}{a+b} f(a-1,b)+ \frac{b}{a+b}f(a+1,b-1) $$ for positive integers $a$ and $b$, with the boundary conditions $f(0,b)=0$ for integers $b>0$ and $f(a,0)=1$ for integers $a\ge0$.

I believe that $f(1,n)$ converges to $1$ as $n$ tends to infinity. However I can't see how to prove it.

Numerically the convergence is slow.

@FedorPetrov's reformulation of the recurrence as a probability question is as follows (lightly edited). Consider an urn with $a$ White and $b$ Black balls. Take a random ball, eat it if it is White and recolor it White (returning to the urn) if it is Black. Then $f(a,b)$ is the probability that when the balls become of the same color, this color is White.

Consider the following recurrence relation in two variables.

$$f(a, b) = \frac{a}{a+b} f(a-1,b)+ \frac{b}{a+b}f(a+1,b-1) $$

with the boundary conditions $f(0,b)=0$ for $b>0$ and $f(a, 0)=1$.

I believe that $f(1,n)$ converges to 1 as $n$ tends to infinity. However I can't see how to prove it.

Numerically the convergence is slow.

Consider the following recurrence relation in two variables: $$f(a, b) = \frac{a}{a+b} f(a-1,b)+ \frac{b}{a+b}f(a+1,b-1) $$ for positive integers $a$ and $b$, with the boundary conditions $f(0,b)=0$ for integers $b>0$ and $f(a,0)=1$ for integers $a\ge0$.

I believe that $f(1,n)$ converges to $1$ as $n$ tends to infinity. However I can't see how to prove it.

Numerically the convergence is slow.