If $a_n = O(n^\alpha)$ and $b_n = o(n^\beta)$, prove that $a_nb_n = o(n^{(\alpha + \beta)})$ and $a_n+b_n = O(\max(n^\alpha, n^\beta))$.

For the part about $a_nb_n = o(n^{(\alpha + \beta)})$, I get that I am supposed to set it up so that

$|a_n| < Mn^\alpha$ for some positive real M for $n \geq n'$

$|b_n| < \epsilon n^\beta$ for all positive reals $\epsilon$ for $n \geq n''$

Then I choose the maximum of n' and n'', and I get

$|a_nb_n| < M\epsilon n^\alpha n^\beta$.

My question is does this finish the proof? If there is a positive real M multiplied by $\epsilon$, does that map to all $\epsilon$? How do you denote this? I think this is some real analysis proof here or I can just say a constant multiplied by epsilon is just epsilon?

For the sum proof, does it become big-O rather than little-O because it cannot be mapped to $\epsilon$ when you add the two parts?

Thank you in advance for the help.

  • $\begingroup$ For the second part, if $\beta > \alpha$ then you do get $o(n^\beta)$, but otherwise, all you can conclude is $O(n^\alpha)$. $\endgroup$ Nov 1, 2020 at 19:16
  • $\begingroup$ Ok thank you, that makes sense. For the first part, is it valid to say something like $\epsilon$ times a constant will result in $\epsilon$? Like this is true from just logic, but I don't know if there's some math theorem I need to state. $\endgroup$
    – Alex
    Nov 1, 2020 at 19:40
  • $\begingroup$ Use the definitions. That's all you need to know. $\endgroup$ Nov 1, 2020 at 19:40

1 Answer 1


For $a_n$ you use correct definition i.e. $a_n \in O(n^\alpha)$: $\exists C>0, \exists N_1 \in \mathbb{N}, \forall n>N_1, |a_n| \leqslant C n^\alpha$.

For little-$o$ for $b_n \in o(n^\beta)$ let's use following definition : $\exists \varepsilon_n, \lim\limits_{n \to \infty}\varepsilon_n=0$ and $ \exists N_2 \in \mathbb{N}$ such that $ \forall n>N_2, b_n=\varepsilon_n n^\beta$.

So, taking $n>\max(N_1,N_2)$ we have $a_n \cdot b_n= \frac{a_n \varepsilon_n}{n^\alpha} \cdot n^{\alpha+\beta} = \phi_n n^{\alpha+\beta}$. We need to prove, that $\phi_n \to 0$: $|\phi_n| = |\frac{a_n \varepsilon_n}{n^\alpha}| \leqslant C |\varepsilon_n|\to 0$.

Hope you'll be able finish second using above.

  • $\begingroup$ Thanks I understand this. I have a small question. When you rewrote like the little-o notation, what happened to like the absolute value part of |bn|. In general, why is there an absolute value part for the definition of |an| to begin with? Furthermore, when you showed $\phi_n$ -> 0, you took the absolute value of it, but your definition of little-o didn't have the $\epsilon$ as an absolute value, so why do you need to consider the absolute value later. Thanks. $\endgroup$
    – Alex
    Nov 1, 2020 at 22:48
  • $\begingroup$ In general definition of little-$o$ do not need absolute value. Estimation for absolute value is result of representation $b_n=\varepsilon_n n^\beta$. As to second question, then $\phi_n \to 0$ is same as $|\phi_n| \to 0$ and in latter we use estimation for absolute value for $a_n$. $\endgroup$
    – zkutch
    Nov 1, 2020 at 23:04

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