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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