Sum of little o Is it true the following?
$$o\left(\frac{1}{n^\alpha}\right)+o\left(\frac{1}{n+1}\right)=o\left(\frac{1}{n+1}\right), \text{ for}\, n\to\infty$$
where $\alpha>1$.
I think yes since I know that $o(x^n)+o(x^m)=o(x^p)$, where $p=\min\{n,m\}$, for $x\to 0$.
So now I have to consider that my $x$ is $\displaystyle\frac{1}{n}$. Am I right?
 A: Yes your idea is right, indeed we have that by definition
$$o\left(\frac{1}{n^\alpha}\right)=\frac{1}{n^\alpha}\cdot \omega_1(n)$$
$$o\left(\frac{1}{n+1}\right)=\frac{1}{n+1}\cdot \omega_2(n)$$
with $\omega_1(n)\to 0$ and $\omega_2(n)\to 0$, then
$$o\left(\frac{1}{n^\alpha}\right)+o(n)=\frac{1}{n^\alpha}\cdot \omega_1(n)+\frac{1}{n+1}\cdot \omega_2(n)=$$
$$=\frac{1}{n+1}\left(\frac{n+1}{n^\alpha}\cdot \omega_1(n)+\omega_2(n)\right)=\frac{1}{n+1}\omega_3(n)$$
with $\omega_3(n)\to 0$, then by definition
$$o\left(\frac{1}{n^\alpha}\right)+o\left(\frac{1}{n+1}\right)=o\left(\frac{1}{n+1}\right)$$
More in general:

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*$o(x^n)=x^n \omega_1(x)$ with $\omega_1(x)\to 0$

*$o(x^m)=x^m \omega_2(x)$ with $\omega_2(x)\to 0$
then assuming wlog $n\le m$
$$o(x^n)+o(x^m)=x^n \omega_1(x)+x^m \omega_2(x)=x^n\left(\omega_1(x)+x^{m-n} \omega_2(x)\right)=x^n \omega_3(x)$$
with $\omega_3(x) \to 0$, then
$$o(x^n)+o(x^m)=o(x^p)$$
with $p=\min\{n,m\}$, which is the rule you are using.
Refer also to the related:

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*Behaviour of little-o in equivalence
