Prove the forthcoming:

g ( n ) ∈ o ( f ( n )) then f ( n ) + g ( n ) ∈ Θ( f ( n )).

Given: g(n) ∈ o(f(n))

Hence, in our condition, coercion total ε>0 we can invent some uniform N such that (+), coercion our functions g(n) and f(n). Hence, coercion n>N, we feel

|g(n)| ≤ ε*|f(n)|, coercion some ε>0, coercion total n>N

Choose a uniform ε < 1 (recall, the over holds coercion total ε > 0),

with accompanied uniform N.

Then the forthcoming holds coercion total n>N

ε(|g(n)| + |f(n)|) ≤ 2|f(n)| ≤ 2(|g(n)| + |f(n)|) ≤ 4*|f(n)| (*)

Stripping detached the left-most disparity in (*) and dividing by 2, we feel:

|f(n)| ≤ |g(n)| + |f(n)| ≤ 2*|f(n)|, n>N (**)

We conceive that this is the very specification Big-Θ notation, as presented in (++), with uniforms k_1 = 1, k_2 = 2 and h(n) = g(n)+f(n). Hence

(**) => g(n) + f(n) is in Θ(f(n))

Ans we feel shown that g(n) ∈ o(f(n)) implies (f(n) + g(n)) ∈ Θ(f(n)).