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Proof

(related to Proposition: Sum of Convergent Real Sequences)

Let \(\epsilon > 0\). Because the real sequences \((a_n)_{n\in\mathbb N}\) and \((b_n)_{n\in\mathbb N}\) are convergent with \(\lim_{n\rightarrow\infty} a_n=a\) and \(\lim_{n\rightarrow\infty} b_n=b\), it follows that

1. there is an \(N(\epsilon)\in\mathbb N\) with \(|a_n - a| < \epsilon\) for all \(n > N(\epsilon)\), and that
2. there is an \(M(\epsilon)\in\mathbb N\) with \(|b_n - b| < \epsilon\) for all \(n > M(\epsilon)\).

Therefore, for all \(n > \max(N(\epsilon),M(\epsilon))\), it follows from triangle property of the absolute value that \[|(a_n + b_n) - (a + b)| \le |a_n - a| + |b_n - b| < \epsilon + \epsilon=2\epsilon.\]

Since \(\epsilon\) can be arbitrarily small chosen, it follows that

\[\lim_{n\rightarrow\infty} (a_n + b_n)=\lim_{n\rightarrow\infty} a_n + \lim_{n\rightarrow\infty} b_n= a + b.\]

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983