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 $\ell_\infty$


Comments

Post a Comment

Popular posts from this blog

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$ Proof. The function $$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$ is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that $$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$ If we make change of variables in the latter integral $t\to 1-t$ we obtain $$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$ On the other hand, using integration by parts, we obtain $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$ Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$ Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$ On the ot
Read more

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds  $$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$ Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$ $\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence $$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$ and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_{n\to\infty}x_n=0.$ $\textbf{
Read more

A problem proposed by prof. Babev

Consider the sequence $$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$ It could be proven that $$\lim_{n\to \infty}n a_n=\frac{1}{2}$$  Now we are going to prove that $$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$ and show how to derive all such limits. $\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as $$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$ $$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$ Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to $$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$ Since for $s\in[0,1]$, $\displaystyle \frac
Read more