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I have the following problem: show that if $\sum_{n=1}^\infty a_n$ is absolutely convergent, there exists an absolutely convergent series $\sum_{n=1}^\infty b_n$ such that $\lim_{n \to \infty} \frac{a_n}{b_n}=0$. Explain why this result shows that there is no "universal" comparison test for testing absolute convergence.

My work: Given that $\sum_{n=1}^\infty a_n$ is absolutely convergent, we know that $\sum_{n=1}^\infty a_n$ is convergent, so $\lim_{n\to \infty} a_n=0$. Would it not just suffice to set $b_k=2a_k$, in order to prevent the denominator from making the fraction large?

What exactly is the idea behind no "universal" comparison test for testing absolute convergence? I think I lack the intuition to understand this.

Thanks in advance!

coffeemath
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DjuroPucar
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    Hi :) The sequence $(b_n) n$ with $b_n=2a_n$ does not fulfill your condition $\lim{n\to\infty}\frac{a_n}{b_n}=0$. Observe, that $\frac{a_n}{b_n}=\frac 12$ for all $n$. – Jochen Dec 26 '22 at 17:06
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    See for example https://math.stackexchange.com/a/452074/42969 – Martin R Dec 26 '22 at 17:09

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This answer is inspired by MartinR's comment.

Let $(c_n)_n$ and $(b_n)_n$ be defined as

$\displaystyle c_n=\sum_{k=n}^\infty |a_k|$ and $b_n=\frac{|a_n|}{\sqrt{c_n}}$.

If we set $\displaystyle s:=\sum_{k=1}^\infty|a_k|$, then $$c_n=s-\sum_{k=1}^{n-1}|a_k|.$$ Observe, that $(c_n)_n$ is a monotonously falling sequence, which converges to zero. Hence, $$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\sqrt{c_n}=0.$$ From $c_n-c_{n+1}=|a_n|$ follows $$ b_n=\frac{1}{\sqrt{c_n}}\cdot(c_n-c_{n+1}), $$ Now, one can approximate it like a Riemann sum by an integral: $$\sum_{n=1}^\infty b_n\leq\int_0^{c_1}\frac{1}{\sqrt x}dx=2\sqrt{c_1}<\infty. $$

Jochen
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