Help proving that $(n+a)^b = \Theta(n^b)$

Question

Please you apologize me by my English.

I don't know how make that: $$(n+a)^b = \Theta(n^b), b > 0$$ I know, I must to find two constants such that: $$ c_{1} n^b \leq (n+a)^b \leq c_{2} n^b $$

I do not know what else to do. I'v tried with the Newton's binomial, but I'm lost.

Answer 1

If $a \geq 0$ then

$$(n+0)^b \leq (n+a)^b \,,$$

thus $c_1=1$ works.

Also, for all $n \geq a$ you have

$$(n+a)^b \leq (2n)^b =2^b n^b \,.$$

Now, fixing

$$c_2 = \max \{ 2^b, \frac{(n+1)^b}{n^b},..., \frac{(n+n-1)^b}{n^b} \}$$

you get the desired inequality.

For $a \leq 0$ you can get the inequlities the other way around, excepting that you'll have an issue if $a$ is a negative integer (what happens if $n=-a$?).

Answer 2

For the upper bound, $(n+a)^b = n^b(1+a/n)^b $. If $n/a > b$, $(1+a/n)^b < (1+1/b)^b < e$, so we can take $c_2 = e$ for large enough $n$.

By taking $n$ large enough compared to $a$ and $b$, we can take any value greater than 1 for $c_2$.