Bounded Sequence and Limit Superior/Inferior with a Scalar

(a) Since (Sn) is bounded, there exists M > 0 such that |Sn| ≤ M for all n. Then for any ε > 0, we have k • lim sup Sn - ε/M ≤ kSn ≤ k • lim sup Sn + ε/M Taking lim sup on both sides and using the properties of lim sup, we get k • lim sup Sn - ε/M ≤ lim sup (ksn) ≤ k • lim sup Sn + ε/M Since ε and M are arbitrary, we can let ε → 0 and M → ∞ to obtain lim sup (ksn) = k • lim sup Sn

(b) Using Exercise 11.8, we have lim inf (-Sn) = - lim sup Sn Then for any ε > 0, we can find N such that -Sn - ε ≤ lim inf (-Sn) ≤ -Sn Adding kSn to both sides and using the properties of lim sup, we get (k-ε) • lim sup Sn ≤ lim inf (ksn) ≤ (k+ε) • lim sup Sn Since ε is arbitrary, we can let ε → 0 to obtain lim inf (ksn) = k • lim sup Sn

(c) If k < 0, then we need to use Exercise 11.8 in a different way: lim sup (-Sn) = - lim inf Sn Then for any ε > 0, we can find N such that -Sn ≤ lim sup (-Sn) + ε ≤ -Sn + ε Multiplying both sides by k and using the properties of lim sup, we get k • lim inf Sn - kε ≤ lim sup (ksn) ≤ k • lim inf Sn Since ε is arbitrary, we can let ε → 0 to obtain lim inf (ksn) = k • lim inf Sn Similarly, we can use Exercise 11.8 to show that lim sup (ksn) = k • lim inf Sn