There is no lowest rate of convergence

 Let $\{a_n\}_{n\ge 1}$ be a sequence of positive numbers such that 

$$\sum_{n=1}^{\infty} a_n$$ is convergent. Then there exists a sequence $\{b_n\}_{n\ge 1}$ with $\displaystyle \lim_{n\to \infty}\frac{b_n}{a_n}=\infty$ and such that 

$$\sum_{n=1}^{\infty} b_n$$ is also convergent.


First proof (a la Calculus I, based on ideas of Zhivko Petrov). Consider $$S_n=\sum_{k=n}^{\infty}a_k.$$ Clearly $S_n\to 0$. Define $$b_n=\sqrt{S_n}-\sqrt{S_{n+1}}.$$

Clearly $b_n>0$ for all $n$ and $b_n\to 0$. Moreover 

$$\sum_{n=1}^\infty b_n=\sqrt{S_1},$$ hence is convergent.

On the other hand $$\lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}\frac{1}{\sqrt{S_n}+\sqrt{S_{n+1}}}=\infty.$$

 

Second proof. (Based on Folland's book) Assume on the contrary that there is lowest decaying sequence, i.e. 

for some sequence $a=\{a_n\}_{n\ge 1}$ with convergent series, holds that for any sequence of positive numbers $\{b_n\}_{n\ge 1}$, the series $$\sum_{n=1}^\infty a_nb_n$$ is convergent if and only if $\{b_n\}_{n\ge 1}$ is bounded (the if part holds for any sequence $\{a_n\}_{n\ge 1}$ with convergent series).

This implies that the linear operator $T:\ell_{\infty}\to\ell_1$ defined by 

$$T(c_1,c_2,\ldots)=(a_1c_1,a_2c_2,\ldots)$$ 

is not only well defined but also surjective ( due to the "only if" part above). Moreover the map is injective, since $a$ has only nonzero terms. The map is clearly bounded (by the norm of $a$), so the bounded inverse mapping theorem implies that the inverse $T^{-1}:\ell_1\to\ell_\infty$ is also bounded, hence continuous operator. Now it remains to observe that the image of $c_{00}$ under $T^{-1}$ is again $c_{00}$. However, $c_{00}$ is dense in $\ell_1$ and its image under continuous map should be dense in $\ell_\infty$, but the image (which is again $c_{00}$) is clearly not dense in $\ell_\infty$


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