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Use the natural filtration ${\mathcal{G}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$. Define the deterministic tail $r_n={\sum }_{j=n}^{\infty }{\delta }_j$ and put $Z_n={\xi }_n+r_n$. This is nonnegative and integrable. Since $r_n={\delta }_n+r_{n+1}$,Thus the deterministic summable drift correction produces a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives a finite limit $Z_{\infty }$ almost surely. Its finiteness is also reflected in the bound $\mathbb{E}{Z}_n\le \mathbb{E}{Z}_1$ and Fatou's lemma, which gives $\mathbb{E}{Z}_{\infty }\le \mathbb{E}{Z}_1<\infty$. The convergence theorem follows from the upcrossing argument: bounded expected negative parts of $-Z$ rule out infinitely many crossings of rational intervals, and the expectation bound excludes an infinite limit.

Since $r_n\to 0$, the required conclusion is $\boxed{{\xi }_n⟶Z_{\infty }\text{ almost surely, with }{Z}_{\infty }<\infty }$. The original sequence need not itself be a supermartingale; subtracting the vanishing deterministic tail recovers it from one that is.