Can we invoke "almost supermartingale" Theorem for deterministic sequences?

Question

Perhaps stupid question.

Question: Can "almost supermartingale" theorem be equally applicable to prove the convergence of some algorithms solving non-random optimization problems?

Attempt for a non-random version of "almost supermartingale" theorem (without any proof):

Let the non-negative sequences be $\{V^k\}$, $\{S^k\}$, and $\{ U^k \}$ for $k=0,1,2,\ldots$. Let $\beta_0, \beta_1$ be non-negative scalars satisfying $\sum_{k=0}^\infty \beta_k < \infty$. Assume $$V^{k+1} \leq \left( 1 + \beta_k \right) V^k - S^k + U^k$$ and $$\sum_{i={\color{red}{0 \text{ or } 1?}}}^\infty U^i < \infty.$$ Then,

1. $V^k \rightarrow V^\infty$
2. $\sum_{k=0}^\infty S^k < \infty$.

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EDIT:

Additionally, let $V^k := \| x^k - x^\star \|_P^2$ where $P$ is a positive semidefinite matrix. Then, using the above theorem, can one say for a subsequence $n_k$ $\lim_{k \rightarrow \infty} x^{n_k} = x^\star$? or do we need more information to prove this?

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For the completeness, I have also copied the "almost supermartingale" theorem from [Theorem 30, page 300].

Theorem: Let $V^k$, $S^k$, and $ U^k$ be $\mathcal{F}_k$-measurable random variables satisfying $V^k \geq 0$, $S^k \geq 0$, and $ U^k\geq 0$ for $k=0,1,2,\ldots$. Let $\beta_0, \beta_1$ be non-negative scalars satisfying $\sum_{k=0}^\infty \beta_k < \infty$. Assume $$\mathbb{E}\left[V^{k+1} \mid \mathcal{F}_k \right] \leq \left( 1 + \beta_k \right) V^k - S^k + U^k$$ and $$\sum_{i={\color{red}{0 \text{ or } 1?}}}^\infty U^i < \infty$$ almost surely. Then,

1. $V^k \rightarrow V^\infty$
2. $\sum_{k=0}^\infty S^k < \infty$.

Answer 1

For $k=0,1,\dots$, let $v_k:=V^k$, $s_k:=S^k$, $u_k:=U^k$, and $b_k:=\beta_k$, so that the $v_k$'s, $s_k$'s, $u_k$'s, and $b_k$'s are nonnegative real numbers such that $\sum_{k=0}^\infty b_k<\infty$, \begin{equation*} \sum_{k=0}^\infty u_k<\infty, \tag{1} \end{equation*} and \begin{equation*} v_{k+1}\le c_k v_k-s_k+u_k \tag{2} \end{equation*} for all $k$, where \begin{equation*} c_k:=1+b_k\ge1, \end{equation*} so that \begin{equation*} \prod_{k=0}^\infty c_k \text{ converges (to a number in $[1,\infty)$).} \tag{3} \end{equation*}

It follows by (2) that $v_{k+1}\le c_k v_k+u_k$ for all $k$ and hence, by induction on $k$, \begin{equation*} v_k\le\Big(v_0+\sum_{j=0}^{k-1}u_j\Big)\prod_{j=0}^{k-1}c_j \tag{4} \end{equation*} and hence \begin{equation*} 0\le v_k\le M:=\Big(v_0+\sum_{j=0}^\infty u_j\Big)\prod_{j=0}^\infty c_j<\infty. \tag{5} \end{equation*}

Similarly to (4), \begin{equation*} v_k\le\Big(v_n+\sum_{j=n}^{k-1}u_j\Big)\prod_{j=n}^{k-1}c_j \tag{6} \end{equation*} for any natural $k$ and $n$ such that $k>n$. Therefore, \begin{equation*} \limsup_{k\to\infty}v_k\le\Big(v_n+\sum_{j=n}^\infty u_j\Big)\prod_{j=n}^\infty c_j \end{equation*} and hence, in view of (1) and (3), \begin{equation*} \limsup_{k\to\infty}v_k\le\big(\liminf_{n\to\infty}v_n+0\big)\times1 = \liminf_{n\to\infty}v_n. \end{equation*} So, in view of (5), there exists \begin{equation*} v_\infty:=\lim_{k\to\infty}v_k\in[0,M|\subset[0,\infty). \tag{7} \end{equation*}

Similarly to (4), for all $k$ \begin{equation*} v_k\le\Big(v_0+\sum_{j=0}^{k-1}u_j\Big)\prod_{j=0}^{k-1}c_j - \sum_{j=0}^{k-1}s_j, \end{equation*} whence \begin{equation*} \sum_{j=0}^{k-1}s_j\le\Big(v_0+\sum_{j=0}^{k-1}u_j\Big)\prod_{j=0}^{k-1}c_j. \end{equation*} Letting now $k\to\infty$ and recalling (1) and (3), we see that \begin{equation*} \sum_{j=0}^\infty s_j<\infty. \tag{8} \end{equation*}

Relations (7) and (8) are what was desired.