Cfrgtkky

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 $a_{n}leqsqrt{frac{a_{n}}{n}}$ $forall$ n $inmathbb{N}$.
Any help would be great. Thank you!

We can get by with assuming $sumlimits_nsqrt{a_n/n}$ 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 $sumlimits_nu_n$ converges.
Since $u_n$ is decreasing,
$$
begin{align}
nu_n
&lesum_{k=1}^nu_ktag1\
&lesum_{k=1}^infty u_ktag2
end{align}
$$
Applying $(2)$ yields
$$
begin{align}
sum_{n=1}^infty nu_n^2
&leleft(sup_{1le nleinfty}nu_nright)sum_{n=1}^infty u_n\
&leleft(sum_{n=1}^infty u_nright)^2tag3
end{align}
$$
Note that $(3)$ is sharp if we consider the sequence
$$
u_n=left{begin{array}{}
1&text{if }n=1\
0&text{if }ngt1
end{array}right.
$$
Furthermore, $(3)$ says that $sumlimits_{n}a_n=sumlimits_{n}nu_n^2$ converges.

Stronger Inequality

Inequality $(3)$ answers the question, but we can get a bit stronger result.
$$
begin{align}
sum_{n=1}^infty nu_n^2
&lesum_{n=1}^inftysum_{k=1}^nu_ku_ntag4\
&=sum_{k=1}^inftysum_{n=k}^infty u_ku_ntag5\
&=frac12left[left(sum_{n=1}^infty u_nright)^2+sum_{n=1}^infty u_n^2right]tag6\
sum_{n=1}^infty(2n-1)u_n^2
&leleft(sum_{n=1}^infty u_nright)^2tag7
end{align}
$$
Explanation:
$(4)$: apply $(1)$
$(5)$: change order of summation
$(6)$: average $(4)$ and $(5)$
$(7)$: subtract $frac12sumlimits_{n=1}^infty u_n^2$ from both sides and double

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

By the Cauchy condensation test, $sum_{n=1}^{infty} sqrt{a_n/n}$ converges if and only if

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

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

Alternative approach: if a sequence ${a_n}_{ngeq 1}$ is such that $sum_{ngeq 1}lambda_n a_n$ is finite for any ${lambda_n}_{ngeq 1}inell^2$, then ${a_n}_{ngeq 1}inell^2$. 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 $sum_{ngeq 1}frac{a_n}{sqrt{n}}=C<+infty$ should be more than enough to ensure that $sum_{ngeq 1}lambda_n a_n$ is finite for any $Lambdainell^2$, since $sum_{ngeq 1}frac{1}{n}$ is divergent.