Limit of a sequence Let $\{u_n\}$ be a sequence of nonnegative numbers satisfying the condition
$$
\tag{1} u_{n+1}\leq (1-\alpha_n)u_n+\beta_n \quad \forall n\in\mathbb{N},
$$
where $\{\alpha_n\}$ and $\{\beta_n\}$ are sequences of real numbers such that
$\tag{2}\displaystyle\lim_{n\rightarrow\infty}\alpha_n=0$ 
$\tag{3}\displaystyle\sum_{n=0}^{\infty}\alpha_n=\infty$
$\tag{4} \displaystyle\sum_{n=0}^{\infty}\beta_n<\infty$
Prove that 
$$\displaystyle\lim_{n\rightarrow\infty}u_n=0$$
 A: Nota: The hypothesis that $\alpha_n\geqslant0$ for every $n$ is obviously missing from the question, and the solution below assumes it.
Preliminary step: Show that one can assume without loss of generality that $\alpha_n\lt1$ for every $n$. From now on, we assume this.
First step: Show by a recursion over $n\geqslant0$ that $u_n\leqslant A_{n-1}u_0+A_{n-1}\sum\limits_{k=0}^{n-1}A_k^{-1}\beta_k$ where $A_{-1}=1$ and, for every $k\geqslant0$, $A_k=\prod\limits_{i=0}^k(1-\alpha_i)$.
Second step: Show that $A_n\to0$ when $n\to\infty$. Hint: $\log A_n\leqslant-\sum\limits_{k=0}^n\alpha_k$.
Third step: Show that, for every $n\geqslant k\geqslant 0$, $u_n\leqslant A_{n-1}u_0+A_{n-1}\sum\limits_{i=0}^{k-1}A_i^{-1}\beta_i+\sum\limits_{i\geqslant k}\beta_i$.
Fourth step: conclude. 
For every $\varepsilon\gt0$, there exists $K_\varepsilon$ such that $\sum\limits_{i\geqslant K_\varepsilon}\beta_i\leqslant\varepsilon$. Then, there exists $M_\varepsilon$ such that $A_{M_\varepsilon-1}\sum\limits_{i=0}^{K_\varepsilon-1}A_i^{-1}\beta_i\leqslant\varepsilon$. Let us choose $N_\varepsilon\geqslant M_\varepsilon$ such that $A_{N_\varepsilon-1}u_0\leqslant\varepsilon$. Then $u_n\leqslant3\varepsilon$ for every $n\geqslant N_\varepsilon$.
A: There must be some missing constraints. If $\alpha_n$ is allowed to be negative, we get the following counterexample. $\smash{\rlap{\phantom{\Bigg(}}}$
Define
$$
u_{n+1}=(1-\alpha_n)u_n+\beta_n\tag{1}
$$
and
$$
A_n=\prod_{k=1}^{n-1}(1-\alpha_k)\tag{2}
$$
By induction, it can be verified that
$$
u_n=A_n\left(u_1+\sum_{k=1}^{n-1}\frac{\beta_k}{A_{k+1}}\right)\tag{3}
$$
For $j\ge1$, define
$$
n_j=\left\{\begin{array}{}
2^{j(j-1)/2}&\text{when }j\text{ is odd}\\
2^{j(j-1)/2+1}&\text{when }j\text{ is even}
\end{array}\right.\tag{4}
$$
and for $n\ge1$,
$$
\alpha_n=\left\{\begin{array}{}
\frac{1}{n+1}&\text{for }n_j\le n< n_{j+1}\text{ when }j\text{ is odd}\\
-\frac1n&\text{for }n_j\le n< n_{j+1}\text{ when }j\text{ is even}
\end{array}\right.\tag{5}
$$
Obviously, $\displaystyle\lim_{n\to\infty}\alpha_n=0$.
Using telescoping products, it is not difficult to show that
$$
\frac{A_{n_{j+1}}}{A_{n_j}}=\left\{\begin{array}{}
\frac{n_j}{n_{j+1}}=2^{-j-1}&\text{when }j\text{ is odd}\\
\frac{n_{j+1}}{n_j}=2^{j-1}&\text{when }j\text{ is even}
\end{array}\right.\tag{6}
$$
Equation $(6)$ yields
$$
A_{n_j}=\left\{\begin{array}{}
2^{-(j-1)/2}&\text{when }j\text{ is odd}\\
2^{-(3j-2)/2}&\text{when }j\text{ is even}
\end{array}\right.\tag{7}
$$
Furthermore, using the standard formula for the partial harmonic series, when $j$ is odd,
$$
\begin{align}
\sum_{n=n_j}^{n_{j+1}-1}\alpha_n
&=\log\left(\frac{n_{j+1}}{n_j}\right)+O\left(\frac{1}{n_j}\right)\\
&=(j+1)\log(2)+O\left(2^{-j(j-1)/2}\right)\tag{8}
\end{align}
$$
and when $j$ is even,
$$
\begin{align}
\sum_{n=n_j}^{n_{j+1}-1}\alpha_n
&=-\log\left(\frac{n_{j+1}}{n_j}\right)+O\left(\frac{1}{n_j}\right)\\
&=-(j-1)\log(2)+O\left(2^{-j(j-1)/2}\right)\tag{9}
\end{align}
$$
Combining $(8)$ and $(9)$ yields
$$
\sum_{n=1}^{n_j-1}\alpha_n=\left\{\begin{array}{}
\frac{j-1}{2}\log(2)+O(1)&\text{when }j\text{ is odd}\\
\frac{3j-2}{2}\log(2)+O(1)&\text{when }j\text{ is even}
\end{array}\right.\tag{10}
$$
Equation $(10)$ says that $\displaystyle\sum_{n=1}^\infty\alpha_n=\infty$.
Define
$$
\beta_n=\left\{\begin{array}{}
2^{-j}&\text{when }n=n_j-1\text{ for }j\text{ even}\\
0&\text{otherwise}
\end{array}\right.\tag{11}
$$
Summing the geometric series yields $\displaystyle\sum_{n=1}^\infty\beta_n=\frac13$.
Using $(3)$, we get
$$
\begin{align}
u_{n_{j+1}}
&=A_{n_{j+1}}\left(u_1+\sum_{k=1}^{n_{j+1}-1}\frac{\beta_k}{A_{k+1}}\right)\\
&\ge\frac{A_{n_{j+1}}}{A_{n_j}}\beta_{n_j-1}\\
&=2^{j-1}\cdot2^{-j}\\
&=\frac12\tag{12}
\end{align}
$$
when $j$ is even. $(12)$ says that $\displaystyle\lim_{n\to\infty}u_n\not=0$.
