In the urn there is one white ball. We do infinite sequence of experiences -- in each step we draw from the urn a ball, we return it into the urn and simultaneously we add black ball. So in the first step we draw white ball, in the second white or black with equal probability etc.
My problems:
$1.$ How to show that with probability $1$ the white ball will be drawn infinitely many times?
$2.$ How to evaluate $\mathbb{E}X_n$ and $Var X_n$, where $X_n$ is an amount of draws to the time of $n$ inclusive in which we have drawn the white ball?
$3.$ Show that the sequence $\{X_n/\log n\}_{n\geq1}$ converges in probability to $1$?
My attempts
$1.$ I have no idea how to do it exactly. I know that probability of drawing white ball in infinitely many experiences is
$P($the white ball will be drawn in every experience$)$=$1\cdot \frac{1}{2}\cdot\frac{1}{3}\cdot \ldots$. Anyway I do not see if it is needed here and how to obtain that the white ball will be drawn infinitely many times with probability $1$.
$2.$My idea was to use the recursion
$\mathbb{E}X_1=1\cdot1=1$,
$\mathbb{E}X_2=1\cdot1\cdot \frac{1}{2}+2\cdot 1 \cdot \frac{1}{2}=\frac12+1=\frac32$,
$\mathbb{E}X_3=1\cdot \frac{1}{2}\cdot\frac23+2(1\cdot\frac12\cdot \frac23+1\cdot\frac12\cdot \frac13 )+3\cdot 1 \cdot \frac{1}{2}\cdot\frac13=\frac13+1+\frac12=\frac{11}{6}$.
Is it a good approach? I do not know if I understood the idea correctly.
$3.$ I have no idea how to begin -- especially how to write $X_n$ down.
Thanks.
The claim that the white ball is drawn infinitely many times is equivalent to saying that there is no positive integer $N$ such that after $N$ draws, the white ball is never drawn again. In other words, pick a number $N$ as large (but finite) as you please. What is the probability that among draws $N+1, N+2, \ldots$, the white ball is never drawn again? It is the product of the individual probabilities that a black ball is drawn at every subsequent draw; i.e., $$\frac{N}{N+1} \cdot \frac{N+1}{N+2} \cdot \frac{N+2}{N+3} \cdot \ldots = \prod_{n=1}^\infty \frac{N+n-1}{N+n}.$$ And what is the limit of this infinite product?
If by $X_n$ you mean the number of times the white ball is drawn in $n$ tries, then your recursion idea is appropriate (but ultimately not necessary). Let $$W_I \sim \operatorname{Bernoulli}(p = 1/i)$$ be equal to $1$ if on draw $i$, the drawn ball is white, and $0$ if black. Then $X_n = \sum_{i=1}^n W_i$, and $$\operatorname{E}[X_n] = \operatorname{E}\left[\sum_{i=1}^{n-1} W_i + W_n\right] = \operatorname{E}[X_{n-1}] + \operatorname{E}[W_n] = \operatorname{E}[X_{n-1}] + \frac{1}{n}.$$ Therefore, $$\operatorname{E}[X_n] = \sum_{i=1}^n \frac{1}{i} = H_n,$$ the $n^{\rm th}$ harmonic number.
The variance is computed similarly, except we have to explicitly state that because the individual draws are independent, the variance of the sum equals the sum of the variances: $$\operatorname{Var}[X_n] \overset{\text{ind}}{=} \operatorname{Var}[X_{n-1}] + \operatorname{Var}[W_n].$$
I leave the third part as an exercise.