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Let $\{a_i\}, i=1, 2, \cdots$ be a nonincreasing sequence of positive numbers, and suppose $\sum_{k=1}^{\infty}\frac{a_k}{\sqrt{k}}<\infty~$, show that $\sum_{k=1}^{\infty}a_k^2<\infty.$

Could you give a direct proof by finding an inequality of the type $\sum_{k=1}^{\infty}a_k^2\leq f<\infty$, where $f$ is some expression related to the already known number $\sum_{k=1}^{\infty}\frac{a_k}{\sqrt{k}}$?

Thanks in advance!


I have solved this problem, but my method is indirect, I have used the uniformly bounded principal and the inequality $lim_{k\to\infty}\frac{1}{\sqrt{k}}\sum_{j=1}^ka_j=0,\forall (a_k)\in l^2$, so I really want to know how to solve the problem by finding a direct inequality.

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Just some intuition: If $\sum a_k /\sqrt{k}$ converges, then $a_k$ goes to zero much faster than $1/ \sqrt{k}$. This implies that $a_k^2$ goes to zero much faster than $\frac{1}{k}$. – JavaMan Dec 18 '12 at 22:34
up vote 7 down vote accepted

Set $b_n=\frac{a_n}{\sqrt n}$, then the result desired is equivalent to $$ \begin{align} \sum_{n=1}^\infty nb_n^2 &=\sum_{n=1}^\infty\sum_{k=1}^n b_n^2\tag{1}\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty b_n^2\tag{2}\\ &\le\sum_{k=1}^\infty b_k\sum_{n=k}^\infty b_n\tag{3}\\ &\le\left(\sum_{n=1}^\infty b_n\right)^2\tag{4} \end{align} $$ where $b_n$ is a monotonically decreasing positive sequence.

Explanation of Steps:

$(1)$ $\displaystyle n=\sum_{k=1}^n1$

$(2)$ change order of summation

$(3)$ $b_k\ge b_n$ for $n\ge k$

$(4)$ $\displaystyle \sum_{n=1}^\infty b_n\ge\sum_{n=k}^\infty b_n$

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This same proof also shows that for $a_n,b_n$ non-negative sequences with $a_n$ non-increasing, $$ \sum_{n=1}^\infty na_nb_n\le\left(\sum_{n=1}^\infty a_n\right)\left(\sum_{n=1}^\infty b_n\right) $$ – robjohn Dec 19 '12 at 8:24
thanks! impressing solution. – ougao Dec 19 '12 at 13:22

Hint: Cauchy condensation test.

As the sequence $\{a_k^2\}$ still is non-increasing, we have to check whether the series $\sum_n 2^na_{2^n}^2$ is convergent. As $\frac{a_k}{\sqrt k}$ is non-increasing, we have by condensation that the series $\sum_n2^n\frac{a_{2^n}}{2^{n/2}}$ is convergent.

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still not know how to proceed.. – ougao Dec 18 '12 at 20:55
Perhaps I am missing something. How does the convergence of $\sum_n2^n\frac{a_{2^n}}{2^{n/2}}$ imply the convergence of $\sum_n 2^na_{2^n}^2$? – robjohn Dec 18 '12 at 23:51
@robjohn $\ell^1\subset \ell^2$. – Davide Giraudo Dec 19 '12 at 12:15
@DavideGiraudo: Doh! Of course. (+1) – robjohn Dec 19 '12 at 14:03

For positive integers $n$, let $u(n) \in \ell^2$ such that $u(n)_j = 1$ for $j \le n$ and $0$ otherwise. Then we can write $a = \sum_{j=1}^\infty b_j u(j)$ where $b_j = a_j - a_{j+1}$. Now let $f(a) = \sum_{j=1}^\infty a_j/\sqrt{j}$. We have $$f(u(n)) = \sum_{j=1}^n \frac{1}{\sqrt{j}} \ge \int_1^{n+1} \frac{dx}{\sqrt{x}} = 2 \sqrt{n+1} - 2 \ge c \sqrt{n} = c \|u(n)\|$$ for some constant $c > 0$ (in fact $c = 2 \sqrt{2}- 2$). Thus $$f(a) = \sum_{j=1}^\infty b_j f(u(j)) \ge c^{-1} \sum_{j=1}^\infty b_j \|u(j)\| \ge c^{-1} \|a\|$$ Actually we should be a bit more careful with limits because $f$ is not a continuous linear functional (and is not defined for all $a$), but an appropriate limiting argument using partial sums should work.

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