Question #8060

Task If (a sub_n) and (b sub_n) are cauchy sequences, show directly that (a sub_n + b sub_n) is also a cauchy sequence

Expert's answer

Question 1. If (an)(a_n) and (bn)(b_n) are Cauchy sequences, show directly that (an+bn)(a_n + b_n) is also a Cauchy sequence.

Solution. Since (an)(a_n) and (bn)(b_n) are Cauchy sequences, for any ε>0\varepsilon > 0 there are N1NN_1 \in \mathbb{N}, such that anam<ε2|a_n - a_m| < \frac{\varepsilon}{2} for all m,n>N1m, n > N_1, and N2NN_2 \in \mathbb{N}, such that bnbm<ε2|b_n - b_m| < \frac{\varepsilon}{2} for all m,n>N2m, n > N_2. Set N=max{N1,N2}N = \max\{N_1, N_2\}. Then for all m,n>Nm, n > N we have


(an+bn)(am+bm)=anam+bnbmanam+bnbm<ε2+ε2=ε.\left| \left(a _ {n} + b _ {n}\right) - \left(a _ {m} + b _ {m}\right) \right| = \left| a _ {n} - a _ {m} + b _ {n} - b _ {m} \right| \leq \left| a _ {n} - a _ {m} \right| + \left| b _ {n} - b _ {m} \right| < \frac {\varepsilon}{2} + \frac {\varepsilon}{2} = \varepsilon .


Thus, (an+bn)(a_{n} + b_{n}) is a Cauchy sequence.

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