= Solution
Use the natural <filtration> $\mathcal G_n=\sigma(\xi_1,\ldots,\xi_n)$. 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 $r_n=\delta_n+r_{n+1}$,
$$
\mathbb E[Z_{n+1}\mid\mathcal G_n]
\leq\xi_n+\delta_n+r_{n+1}=Z_n.
$$
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 $\mathbb EZ_n\leq\mathbb EZ_1$ and <Fatou's lemma>, which gives $\mathbb EZ_\infty\leq\mathbb EZ_1<\infty$. 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 $r_n\to0$, the required conclusion is $\boxed{\xi_n\longrightarrow Z_\infty\text{ almost surely, with }Z_\infty<\infty}$. The original sequence need not itself be a <supermartingale>; subtracting the vanishing deterministic tail recovers it from one that is.
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