If we have a sequence $b_n$ whose limit is 0 as n tends to infinity, we can find another sequence $a_n$ such that its limit is also 0 but $b_n/a_n$ again has limit 0 (taking $a_n = \sqrt{|b_n|}$ and taking care case where $b_n$ is eventually 0 separately.

However, suppose we make a stricter condition: not only must the limit of $b_n$ be 0, but $\sum{b_n}$ must have a finite value. Similarly, $\sum{a_n}$ must also have a finite value. Can we again find such a sequence? Simply taking square roots will not work, since $\sum{\frac{1}{n^2}}$ has a finite value but $\sum{\frac{1}{n}}$ does not.

  • $\begingroup$ Do you mean if $a_n\rightarrow0$, $b_n\rightarrow0$, $\sum a_n$ converges, and $\sum b_n$ converges, then $a_n/b_n\rightarrow0$? $\endgroup$ – Jika Jun 26 '14 at 16:09
  • $\begingroup$ With your notation, what I want is that given such an $a_n$ I want to find a $b_n$ so that these conditions are satisfied. $\endgroup$ – Wonder Jun 26 '14 at 16:16

Let $b_n>0$ such that $\displaystyle \sum b_n<+\infty$. Put $\displaystyle R_n=\sum_{k\geq n} b_k$. Then $R_n$ goes to $0$ as $n$ goes to infinity. Let $\displaystyle a_n=\frac{b_n}{\sqrt{R_{n}}}$. We have $\displaystyle \frac{b_n}{a_n}=\sqrt{R_{n}} \to 0$.

As $b_n=R_n-R_{n+1}$ we have

$$a_n=\frac{R_n-R_{n+1}}{\sqrt{R_n}}\leq \int_{R_{n+1}}^{R_n}\frac{dt}{\sqrt{t}}$$ Hence

$$\sum a_n\leq \int_0^{R_1}\frac{dt}{\sqrt{t}}<+\infty$$

And we can re-apply the procedure infinitely.

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  • $\begingroup$ This is great, thanks! $\endgroup$ – Wonder Jun 26 '14 at 16:18

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