<s>[INST] Let $a_n$ be a sequence defined by: $a_n = \min(m)$ $s.t.$ $\sum_{i = 1}^m \sqrt i > n^2$. Does $\sum \frac{1}{a_n}$ converges? I am trying to show convergence (Or divergence) of the following sum:

Let $a_n$ be a sequence defined by:
$a_n = \min(m)$ $s.t.$ $\sum_{i = 1}^m \sqrt i > n^2$.
Does $\sum \frac{1}{a_n}$ converges?

If we define $b_n = \min(m)$ $s.t.$ $\sum_{i = 1}^m \ i > n^2$, We get $b_n > n$, since  $\sum_{i = 1}^ni = \frac{n(n+1)}{2} < n^2$, So $a_n > n$ for sure.
But im not sure how evaluate  $a_n$ any further.
Any hints will be appericiated.
 [/INST] For $a>b>0$ we have
$$a^3-b^3=(a-b)(a^2+ab+b^2)\ge a^2(a-b)$$ $$a-b={a^2-b^2\over a+b}\ge {a^2-b^2\over 2a}$$ Hence for $a=i^{1/2}$ and $b=(i-1)^{1/2}$ we get
$$i^{3/2}-(i-1)^{3/2}\ge i [i^{1/2}-(i-1)^{1/2}]\ge {1\over 2} i^{1/2}$$ Adding the terms gives
$$\displaystyle 2m^{3/2}\ge \sum_{i=1}^m i^{1/2}$$ If the sum is greater than $n^2,$ then $2m^{3/2}\ge n^2, $ which implies $m\ge 2^{-1}n^{4/3}. $ Hence $$a_n\ge {1\over 2}n^{4/3}$$
Remark The estimate is optimal concerning the exponent $4/3,$ that means $a_n\le cn^{4/3}.$ To this end it suffices to apply the inequalities (for $a>b>0$)$$a^3-b^3=(a-b)(a^2+ab+b^2)\le 3a^2(a-b)$$ $$a-b={a^2-b^2\over a+b}\le {a^2-b^2\over a}$$
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