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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?

Lalit Tolani
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Nik
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  • Is there any particular reason for writing $o({1\over n+1})$ instead of $o({1\over n})$? – Barry Cipra Aug 30 '21 at 08:44
  • Do you mean I can write: $o\left(\frac{1}{n^\alpha}\right)+o\left(\frac{1}{n+1}\right)=o\left(\frac{1}{n^\alpha}\right)+o\left(\frac{1}{n}\right)$? – Nik Aug 30 '21 at 08:50
  • I don't see why not, unless $o({1\over n+1})=o({1\over n})$ is, in effect, what you're supposed to prove. – Barry Cipra Aug 30 '21 at 09:59

1 Answers1

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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:

  • $o(x^n)=x^n \omega_1(x)$ 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:

Lorenzo B.
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user
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