Telescoping Sum of expectations: limsup exists but limes not necessarily

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Let $X_t$ for $t \in \{0, 1, \dotsc, \}$ be a sequence of non-negative integer-valued random variables. Suppose that $$\mathbb{E}[X_t - X_{t+1} \mid X_t>0 ] \leq c \quad \text{ for some constant } c \text{, for all } t \geq 0$$ and for all possible outcomes $\omega \in \Omega$.

Moreover assume that

$$\liminf_{t \rightarrow \infty} \mathbb{E}[X_t] = 0.$$

Can we show that

$$\sum_{t=0}^{\infty} \left(\mathbb{E}[X_t]- \mathbb{E}[X_{t+1}]\right)\geq \mathbb{E}[X_0]?$$

Concrete Example: Random Declines. Let $X_0=n$ for $n \rightarrow \infty$. Hence we have a deterministic starting value. Then choose $X_{t+1}$ uniformly at random from the set $\{0, \dotsc, \lfloor e X_t \rfloor\}$ for all $t \geq 0$. Can we say that $\sum_{t=0}^{\infty} (\mathbb{E}[X_t]-\mathbb{E}[X_{t+1}])\geq \mathbb{E}[X_0]= n$?

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Think about the sequence of independent random variables $(X_t)_{t \in \mathbb{N}}$ where \begin{align} X_t & \sim Unif((1,2)) &~\text{if}~& t=2k \\ X_t & \sim Unif((0,\frac{1}{t})) &~\text{if}~& t=2k + 1 \end{align} for any $k \in \mathbb{N}$. Then your sum does not converge.

If you want to prove that sum $$\sum_{n=1}^{\infty}a_n$$ converge (so you can define something using it) you have to know (prove) that
$$\lim_{n \to \infty} a_n = 0.$$ Otherwise the sum is not defined.

Edit: Since OP put another assumption I have to create another counterexample: think about the sequence of independent random variables $(X_t)_{t \in \mathbb{N}}$ such that \begin{align} X_t & \sim Y &~\text{if}~& t=2k \\ X_t & \sim \frac{1}{t}.Y &~\text{if}~& t=2k + 1 \end{align} where $Y$ has Bernouli distribution with $p=0.5$ Again your sum does not converge.

Existence of limit of $a_n$:

If $$\sum_{n=1}^{\infty}a_n = \lim_{k \to \infty} \sum_{n=1}^{k}a_n $$ exists, than for all $m >1$ $$|a_m| \leq \left| \sum_{n=1}^{m}a_n - \sum_{n=1}^{m-1}a_n \right|. $$ Since you assume that the infinite sum converge, the right hand side of the previous inequality goes to $0$ as $m \to \infty$. So the $$\lim_{m \to \infty} |a_m| = 0.$$

Edit 2:

Counterexample no. 3: define sequence of independent random variables $(X_t)_{t \in \mathbb{N}}$, such that every $X_t$ has Bernoulli distribution with parameter $p_t$. Define these parameters as follows \begin{align} p_t &=\frac{1}{2} &~\text{if}~& t=2k, \\ p_t &= \frac{1}{t} &~\text{if}~& t=2k + 1. \end{align} Again the desired sum does not converge.