# Suppose $\sum_{n=1}^{\infty}\sqrt{{a_{n}}/{n}}$ is convergent. Prove that $\sum_{n=1}^{\infty}a_{n}$ is also convergent.

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!

• It is notable that $\sqrt{a_n/n}$ is the geometric mean of $a_n$ and $\frac 1n$ – Ben Grossmann Jan 2 '19 at 3:58
• Why the downvote and the vote to close? – JavaMan Jan 2 '19 at 4:19

We can get by with assuming $$\sum\limits_n\sqrt{a_n/n}$$ converges, and the weaker condition that $$a_n/n$$ is decreasing.

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 $$u_n=\left\{\begin{array}{} 1&\text{if }1\le n\le N\\ 0&\text{if }n\gt N \end{array}\right.$$

• 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 – DanielWainfleet Jan 2 '19 at 10:41

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 $$\sum_n b_n^2$$.

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 $$0 and $$\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.

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