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Let $\{a_{n}\}$ be a decreasing sequence of non-negative real numbers. Suppose $\sum_{n=1}^{\infty}\sqrt{\frac{a_{n}}{n}}$ is convergent. Prove that $\sum_{n=1}^{\infty}a_{n}$ is also convergent. My thought is to use the direct comparison test but I think I'm struggling with showing that $a_{n}\leq\sqrt{\frac{a_{n}}{n}}$ $\forall$ n $\in\mathbb{N}$. Any help would be great. Thank you!

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    $\begingroup$ It is notable that $\sqrt{a_n/n}$ is the geometric mean of $a_n$ and $\frac 1n$ $\endgroup$ – Ben Grossmann Jan 2 '19 at 3:58
  • $\begingroup$ Why the downvote and the vote to close? $\endgroup$ – JavaMan Jan 2 '19 at 4:19
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We can get by with assuming $\sum\limits_n\sqrt{a_n/n}$ converges, and the weaker condition that $a_n/n$ is decreasing.


Answering the Question

Let $u_n=\sqrt{a_n/n}$, then $u_n$ is decreasing and $\sum\limits_nu_n$ converges. Since $u_n$ is decreasing, $$ \begin{align} nu_n &\le\sum_{k=1}^nu_k\tag1\\ &\le\sum_{k=1}^\infty u_k\tag2 \end{align} $$ Applying $(2)$ yields $$ \begin{align} \sum_{n=1}^\infty nu_n^2 &\le\left(\sup_{1\le n\le\infty}nu_n\right)\sum_{n=1}^\infty u_n\\ &\le\left(\sum_{n=1}^\infty u_n\right)^2\tag3 \end{align} $$ Note that $(3)$ is sharp if we consider the sequence $$ u_n=\left\{\begin{array}{} 1&\text{if }n=1\\ 0&\text{if }n\gt1 \end{array}\right. $$ Furthermore, $(3)$ says that $\sum\limits_{n}a_n=\sum\limits_{n}nu_n^2$ converges.


Stronger Inequality

Inequality $(3)$ answers the question, but we can get a bit stronger result. $$ \begin{align} \sum_{n=1}^\infty nu_n^2 &\le\sum_{n=1}^\infty\sum_{k=1}^nu_ku_n\tag4\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty u_ku_n\tag5\\ &=\frac12\left[\left(\sum_{n=1}^\infty u_n\right)^2+\sum_{n=1}^\infty u_n^2\right]\tag6\\ \sum_{n=1}^\infty(2n-1)u_n^2 &\le\left(\sum_{n=1}^\infty u_n\right)^2\tag7 \end{align} $$ Explanation:
$(4)$: apply $(1)$
$(5)$: change order of summation
$(6)$: average $(4)$ and $(5)$
$(7)$: subtract $\frac12\sum\limits_{n=1}^\infty u_n^2$ from both sides and double

Note that $(7)$ is sharp if we consider any of the sequences $$ u_n=\left\{\begin{array}{} 1&\text{if }1\le n\le N\\ 0&\text{if }n\gt N \end{array}\right. $$

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  • $\begingroup$ Letting $v_n= nu_n^2=(nu_n)u_n,$ we have, in the non-trivial case where every $u_n>0$, that $\lim_{n\to \infty}v_n/u_n=0,$ so if $\sum u_n$ converges then $\sum v_n$ does too. ................+1 $\endgroup$ – DanielWainfleet Jan 2 '19 at 10:41
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By the Cauchy condensation test, $\sum_{n=1}^{\infty} \sqrt{a_n/n}$ converges if and only if

$$ \sum_{n=1}^{\infty} 2^n\sqrt{a_{2^n}/2^n} = \sum_{n=1}^{\infty} \sqrt{2^n a_{2^n}} $$

converges. Likewise, $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=1}^{\infty} 2^n a_{2^n}$ converges. Now the conclusion follows from the observation that, if $b_n \geq 0$ and $\sum_n b_n$ converges, then so does $\sum_n b_n^2$.

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Let $b_n=\sqrt{a_n}$. Then $\{b_n\}_{n\geq 1}$ is a decreasing sequence of positive real numbers and we want to show that $$ \sum_{n\geq 1}\frac{b_n}{\sqrt{n}}<+\infty\quad\Longrightarrow\quad \sum_{n\geq 1}b_n^2 < +\infty.$$ By the Cauchy-Schwarz inequality disguised as Titu's lemma we have $$ \sum_{n=N+1}^{2N}\frac{b_n}{\sqrt{n}} \geq \frac{\left(\sqrt{b_{N+1}}+\ldots+\sqrt{b_{2N}}\right)^2}{\sqrt{N+1}+\ldots+\sqrt{2N}}\geq \frac{N^2 b_{2N}}{\frac{2}{3}(2\sqrt{2}-1)N\sqrt{N}}\geq \frac{4}{5}\sqrt{N}\,b_{2N} $$ hence $$ \sum_{k\geq 0}2^{k/2} b_{2^k} $$ is convergent and $b_{2^k}=o(2^{-k/2})$. On the other hand $$\begin{eqnarray*} \sum_{n=N+1}^{2N}b_n^2 &=& \sum_{n=N+1}^{2N}\sqrt{n}b_n\cdot\frac{b_n}{\sqrt{n}}\\&=&\sqrt{2N} b_{2N}\sum_{n=N+1}^{2N}\frac{b_n}{\sqrt{n}}+\sum_{n=N+1}^{2N-1}\left(\frac{b_n}{\sqrt{n}}-\frac{b_{n+1}}{\sqrt{n+1}}\right)\sum_{m=N+1}^{n}\frac{b_n}{\sqrt{n}}\end{eqnarray*} $$ by summation by parts, and assuming that $b_n$ is normalized in such a way that $\sum_{n\geq 1}\frac{b_n}{\sqrt{n}}=1$, the RHS is bounded by $$\begin{eqnarray*}&&\sum_{n=N+1}^{2N-1}\frac{b_n}{\sqrt{n}}\left[\sqrt{2N} b_{2N}+\sum_{n=N+1}^{2N-1}\left(\frac{b_n}{\sqrt{n}}-\frac{b_{n+1}}{\sqrt{n+1}}\right)\right]\\&=&\sum_{n=N+1}^{2N-1}\frac{b_n}{\sqrt{n}}\left[\sqrt{2N} b_{2N}+\frac{b_{N+1}}{\sqrt{N+1}}-\frac{b_{2N}}{\sqrt{2N}}\right]\\&\leq& \sum_{n=N+1}^{2N-1}\frac{b_n}{\sqrt{n}}\left[\sqrt{2N} b_{2N}+\frac{5\sqrt{2}}{4N}\right]\\ &\leq&\frac{5\sqrt{2}}{4}\cdot\frac{N+2}{N+1}\sum_{n=N+1}^{2N}\frac{b_n}{\sqrt{n}}.\end{eqnarray*} $$ This detour gives a quantitative improvement of the other proofs: by summing on $N=2^k$ with $k\in\mathbb{N}$, then getting rid of the normalization assumption, we get $$\boxed{\sum_{n\geq 1}b_n^2 \leq b_1^2+\color{blue}{\frac{15\sqrt{2}}{8}}\left(\sum_{n\geq 1}\frac{b_n}{\sqrt{n}}\right)^2} $$ and we may start wondering about the optimal constant that can replace $\frac{15\sqrt{2}}{8}$, like in Hardy's inequality. In this regard it makes sense to replace the short sums $\sum_{n=N+1}^{2N}$ with $\sum_{n=N+1}^{AN}$ and optimize on $A$. By considering $b_n=\frac{1}{n^{1/2+\varepsilon}}$, such that $\sum_{n\geq 1}\frac{b_n}{\sqrt{n}}$ is just barely convergent, we have that the blue constant cannot be improved beyond $\frac{7}{74}$.


Alternative approach: if a sequence $\{a_n\}_{n\geq 1}$ is such that $\sum_{n\geq 1}\lambda_n a_n$ is finite for any $\{\lambda_n\}_{n\geq 1}\in\ell^2$, then $\{a_n\}_{n\geq 1}\in\ell^2$. This is a consequence of the Banach-Steinhaus theorem, which can be proved independently by summation by parts (see page 150 of my notes). The constraints $0<a_{n+1}<a_n$ and $\sum_{n\geq 1}\frac{a_n}{\sqrt{n}}=C<+\infty$ should be more than enough to ensure that $\sum_{n\geq 1}\lambda_n a_n$ is finite for any $\Lambda\in\ell^2$, since $\sum_{n\geq 1}\frac{1}{n}$ is divergent.

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  • $\begingroup$ This question is possibly related to your alternative approach. $\endgroup$ – robjohn Jan 3 '19 at 21:10

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