$u_n = 1/n^2$ if $n \in [\![k^2, (k+1)^2]\!] , k \in 2\mathbb{N}+1$ and $u_n = 1/([\sqrt{n}]^4+n-[\sqrt{n}]^2)$, does $\sum u_n$ converge

Question

Let $(u_n)_{n \in \mathbb{N}^*}$ such that :

$u_n = 1/n^2$ if $n \in [\![k^2, (k+1)^2]\!] , k \in 2\mathbb{N}+1$ and $u_n = 1/([\sqrt{n}]^4+n-[\sqrt{n}]^2)$ if $n \in [\![v^2, (v+1)^2 [\![, v \in 2\mathbb{N}$

Hence we have : $u_1 =1, u_2 = 1/4, u_3 = 1/9, u_4 = 1/16, u_5 =1/17, u_6 = 1/18, ...$

I would like to know if : $\sum u_n$ converge.

What I've note so far is that : $u_n$ is strictly decreasing, and has limit $0$.

$\pi^2/6 \leq \sum u_n \leq H_n$, but it's hard for me to really see how to advanced.

Answer 1

Want $u_n = 1/([\sqrt{n}]^4+n-[\sqrt{n}]^2) $

If $(2k)^2 \le n \lt (2k+1)^2 $, let $j = n-(2k)^2 $. Then $0 \le j \le (2k+1)^2-(2k)^2-1 =4k $ and $\sqrt{n} \lt 2k+1 $ so $k \gt (\sqrt{n}-1)/2 $.

$[\sqrt{n}]^4+n-[\sqrt{n}]^2 =(2k)^4+((2k)^2+j)-(2k)^2 =(2k)^4+j \gt ((\sqrt{n}-1)/2)^4 \gt n^2/256 $ for $n \gt 4$.

So the sum of the reciprocals of those $n$ is less than $256\sum\frac1{n^2}$ which converges.

Answer 2

$[\sqrt n\;]^4+n- [\sqrt n]^2\;\geq\;$ $ [\sqrt n\;]^4+n- (\sqrt n\;)^2=\;$ $\;[\sqrt n\;]^4 \geq\;$ $\;(-1+\sqrt n\;)^4 .$

So for all but finitely many $n$ we have $0<u_n\leq 42/n^2.$