With asymptotic formulas for partial sum Conjecture
If $(u_n)$ is a decreasing sequence of strictly positive numbers and if $$\sum_{k=1}^n u_k\sim \sqrt n$$ Show  that $u_n\sim \frac 1{2 \sqrt n}$
example $u_n=\sqrt{n+1}-\sqrt{n}$
I can't find a counter example
 A: It is some sort  of converse of Cesaro, but with a bit of a trick. Define
$$
S_n = \sum_{k=1}^n u_k,
$$
take $\epsilon \in (0,1)$, and let $a_n = \lfloor(1+\epsilon) n \rfloor$. By monotonicity, we have
$$
u_{n+1} + u_{n+2} + \cdots + u_{n+a_n} \leq (a_n-n) u_n \leq u_n + u_{n-1} + \cdots + u_{n-(a_n-n)+1},
$$
that is
$$
S_{n+a_n} - S_n \leq (a_n - n) u_n \leq S_n - S_{2n - a_n}.
$$
Dividing by $\sqrt{n}$ yields
$$
\sqrt{\frac{n+a_n}{n}} \frac{1}{\sqrt{n+a_n}}S_{n+a_n} - \frac{1}{\sqrt{n}}S_n \leq \frac{a_n - n}{n} \sqrt{n} u_n \leq \frac{1}{\sqrt{n}}S_n - \sqrt{\frac{2n-a_n}{n}} \frac{1}{\sqrt{2n-a_n}} S_{2n - a_n}.
$$
Taking the $\liminf$ on the left and using the assumption and the definition of $(a_n)$ yields
$$
\sqrt{1+\epsilon} - 1 \leq \epsilon \liminf \sqrt{n} u_n.
$$
Dividing by $\epsilon$ and taking the limit as $\epsilon \to 0$ gives
$$
\frac12 \leq \liminf \sqrt{n} u_n.
$$
Similarly, taking the $\limsup$ on the right yields
$$
\epsilon \limsup \sqrt{n} u_n \leq 1-\sqrt{1-\epsilon},
$$
so in the same way
$$
\limsup \sqrt{n} u_n \leq \frac12.
$$
This tells you exactly that
$$
\lim_{n \to + \infty} \sqrt{n}{u_n} = \frac12,
$$
that is $u_n \sim \frac{1}{2 \sqrt{n}}$.
