Proving if g(n) is o(f(n)), then f(n) + g(n) is Theta(f(n))

Question

So I'm struggling with proving (or disproving) the above question. I feel like it is true, but I'm not sure how to show it.

Again, the question is if g(n) is o(f(n)), then f(n) + g(n) is Theta(f(n))

Note, that is a little-o , not a big-o!!!

So far, I've managed to (easily) show that:

g(n) = o(f(n)) -> g(n) < c*f(n)

Then g(n) + f(n) < (c+1)*f(n) -> (g(n) + f(n)) = O (f(n))

However, for showing Big Omega, I'm not sure what to do there.

Am I going about this right?

EDIT: Everyone provided great help, but I could only mark one. THANK YOU.

Answer 1

One option would be to take the limit of (f(n) + g(n)) / f(n) as n tends toward infinity. If this converges to a finite, nonzero value, then f(n) + g(n) = Θ(f(n)).

Assuming that f(n) is nonzero for sufficiently large n, the above ratio, in the limit, is

(f(n) + g(n)) / f(n)

= f(n) / f(n) + g(n) / f(n)

= 1 + g(n) / f(n).

Therefore, taking the limit as n goes to infinity, the above expression converges to 1 because the ratio goes to zero (this is what it means for g(n) to be o(f(n)).

Answer 2

So far so good.

For the next step, recall that in the best case, 0 <= g(n) ; this should get you a lower bound on g(n) + f(n) .

Answer 3

Before we begin, lets first state what little-o and Big-Theta notations means:

Little-o notation

Formally, that g(n) = o(f(n)) (or g(n) ∈ o(f(n)) ) holds for sufficiently large n means that for every positive constant ε there exists a constant N such that

 |g(n)| ≤ ε*|f(n)|, for all n > N (+) 

From https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation .

Big-Θ notation

h(n) = Θ(f(n)) means there exists positive constants k_1 , k_2 and N , such that k_1 · |f(n)| and k_2 · |f(n)| is an upper bound and lower bound on on |h(n)| , respectively, for n > N , ie

 k_1 · |f(n)| ≤ |h(n)| ≤ k_2 · |f(n)|, for all n > N (++) 

From https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-big-theta-notation .

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Given: g(n) ∈ o(f(n))

Hence, in our case, for every ε>0 we can find some constant N such that (+) , for our functions g(n) and f(n) . Hence, for n>N , we have

|g(n)| ≤ ε*|f(n)|, for some ε>0, for all n>N

Choose a constant ε < 1 (recall, the above holds for all ε > 0), 
with accompanied constant N. 
Then the following holds for all n>N

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

Stripping away the left-most inequality in (*) and dividing by 2, we have:

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

We see that this is the very definition Big-Θ notation, as presented in (++) , with constants 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 have shown that g(n) ∈ o(f(n)) implies (g(n) + f(n)) ∈ Θ(f(n)) .