# 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 VerĂ¤nderlichen", Vieweg Studium, 1983