Let \((a_n)_{n\ge1}\) and \((b_n)_{n\ge1}\) be real sequences. Suppose \[ 0\le a_{n+1}\le a_n+b_n \] for every \(n\ge1\), and suppose \(\sum_{n=1}^\infty b_n\) converges. Prove that \(\lim_{n\to\infty}a_n\) exists and is finite.
Solution
1. For every \(\epsilon>0\), there exists \(N\) such that for all\(m>n\ge N\),\[ \left|\sum_{j=n}^{m-1}b_j\right|<\epsilon. \]
The series \(\sum_{j=1}^{\infty}b_j\) converges, so its sequence of partial sums is Cauchy. The displayed estimate is exactly the Cauchy criterion for the tail from \(n\) through \(m-1\).
2. If \(m>n\), then\[ a_m \le a_n+\sum_{j=n}^{m-1}b_j. \]
Iterate the inequalities \[ a_{j+1}\le a_j+b_j \] for \(j=n,n+1,\ldots,m-1\). The intermediate \(a_j\) terms telescope, giving the displayed bound.
3. For every \(\epsilon>0\), if \(N\) is as in step 1, then\[ a_m<a_n+\epsilon \]whenever \(m>n\ge N\).
Combine steps 1 and 2: \[ a_m \le a_n+\sum_{j=n}^{m-1}b_j \le a_n+\left|\sum_{j=n}^{m-1}b_j\right| < a_n+\epsilon. \]
4. The sequence \((a_n)\) is bounded.
The hypothesis gives \(a_n\ge0\) for every \(n\ge2\), so the sequence is bounded below. Fix \(\epsilon=1\) in step 3 and the corresponding \(N\). Then every \(m>N\) satisfies \[ a_m<a_N+1. \] Together with the finitely many terms \(a_1,\ldots,a_N\), this gives an upper bound.
5. One has\[ \limsup_{n\to\infty}a_n \le \liminf_{n\to\infty}a_n+\epsilon \]for every \(\epsilon>0\).
1. For every \(n\ge N\),step 3 gives\[ \sup_{m>n}a_m\le a_n+\epsilon. \]By step 4, both \(\limsup a_n\) and \(\liminf a_n\) are finite.Taking \(\liminf\) as \(n\to\infty\) on the right and using\[ \lim_{n\to\infty}\sup_{m>n}a_m = \limsup_{n\to\infty}a_n \]gives the claimed inequality.
6. The limit of \((a_n)\) exists and is finite.
Letting \(\epsilon\downarrow0\) in step 5 gives \[ \limsup_{n\to\infty}a_n \le \liminf_{n\to\infty}a_n. \] The reverse inequality always holds, so the two are equal. By step 4 their common value is finite. Hence \[ \boxed{\lim_{n\to\infty}a_n\text{ exists and is finite}}. \]
7. Q.E.D.
Step 6 proves the required conclusion.