Proving that a sequence [$a_n = \frac{v_n}{n},\; n \ge 2$] converges.

Question

For $n \ge 2$ we define numbers $a_n$ as follows:

$\quad \text{Set } u_n = [\frac{n \times n}{2}] \quad \text{ ([.] is the floor function)}$

$\quad \text{Set } v_n = \text{ the greatest positive integer with } v_n^2 \le u_n$

$\quad \text{Set } a_n = \frac{v_n}{n}$

Question: Is there a proof that the sequence $(a_n)$ converges?

My Work

This Python program is all I have...

import math

def getY(daInt):
    y = 1
    finis = -1
    while finis < 0:
        if y * y > daInt:
            return y - 1
        y = y + 1

for x in range(2, 101):
    myInt = math.floor(x * x/2)
    myY = getY(myInt)
    print(x, myInt, myY, myY/x)

Program Output:

2 2 1 0.5
3 4 2 0.6666666666666666
4 8 2 0.5
5 12 3 0.6
6 18 4 0.6666666666666666
7 24 4 0.5714285714285714
8 32 5 0.625
9 40 6 0.6666666666666666
10 50 7 0.7
11 60 7 0.6363636363636364
12 72 8 0.6666666666666666
13 84 9 0.6923076923076923
14 98 9 0.6428571428571429
15 112 10 0.6666666666666666
16 128 11 0.6875
17 144 12 0.7058823529411765
18 162 12 0.6666666666666666
19 180 13 0.6842105263157895
20 200 14 0.7
21 220 14 0.6666666666666666
22 242 15 0.6818181818181818
23 264 16 0.6956521739130435
24 288 16 0.6666666666666666
25 312 17 0.68
26 338 18 0.6923076923076923
27 364 19 0.7037037037037037
28 392 19 0.6785714285714286
29 420 20 0.6896551724137931
30 450 21 0.7
31 480 21 0.6774193548387096
32 512 22 0.6875
33 544 23 0.696969696969697
34 578 24 0.7058823529411765
35 612 24 0.6857142857142857
36 648 25 0.6944444444444444
37 684 26 0.7027027027027027
38 722 26 0.6842105263157895
39 760 27 0.6923076923076923
40 800 28 0.7
41 840 28 0.6829268292682927
42 882 29 0.6904761904761905
43 924 30 0.6976744186046512
44 968 31 0.7045454545454546
45 1012 31 0.6888888888888889
46 1058 32 0.6956521739130435
47 1104 33 0.7021276595744681
48 1152 33 0.6875
49 1200 34 0.6938775510204082
50 1250 35 0.7
51 1300 36 0.7058823529411765
52 1352 36 0.6923076923076923
53 1404 37 0.6981132075471698
54 1458 38 0.7037037037037037
55 1512 38 0.6909090909090909
56 1568 39 0.6964285714285714
57 1624 40 0.7017543859649122
58 1682 41 0.7068965517241379
59 1740 41 0.6949152542372882
60 1800 42 0.7
61 1860 43 0.7049180327868853
62 1922 43 0.6935483870967742
63 1984 44 0.6984126984126984
64 2048 45 0.703125
65 2112 45 0.6923076923076923
66 2178 46 0.696969696969697
67 2244 47 0.7014925373134329
68 2312 48 0.7058823529411765
69 2380 48 0.6956521739130435
70 2450 49 0.7
71 2520 50 0.704225352112676
72 2592 50 0.6944444444444444
73 2664 51 0.6986301369863014
74 2738 52 0.7027027027027027
75 2812 53 0.7066666666666667
76 2888 53 0.6973684210526315
77 2964 54 0.7012987012987013
78 3042 55 0.7051282051282052
79 3120 55 0.6962025316455697
80 3200 56 0.7
81 3280 57 0.7037037037037037
82 3362 57 0.6951219512195121
83 3444 58 0.6987951807228916
84 3528 59 0.7023809523809523
85 3612 60 0.7058823529411765
86 3698 60 0.6976744186046512
87 3784 61 0.7011494252873564
88 3872 62 0.7045454545454546
89 3960 62 0.6966292134831461
90 4050 63 0.7
91 4140 64 0.7032967032967034
92 4232 65 0.7065217391304348
93 4324 65 0.6989247311827957
94 4418 66 0.7021276595744681
95 4512 67 0.7052631578947368
96 4608 67 0.6979166666666666
97 4704 68 0.7010309278350515
98 4802 69 0.7040816326530612
99 4900 70 0.7070707070707071
100 5000 70 0.7

Answer 1

As indicated in the comments, if $n=2m$ then $u_{2m}=\left[\frac{2m\cdot 2m}{2}\right]=2m^2$. If $n=2m+1$ then $u_{2m+1}=\left[\frac{(2m+1)\cdot (2m+1)}{2}\right]=2m^2+2m$.

From the above, we have $$v_{2m}^2\leq u_{2m} < (v_{2m}+1)^2 \Rightarrow 0\leq v_{2m}\leq\sqrt{u_{2m}}<v_{2m}+1$$ $$v_{2m+1}^2\leq u_{2m+1}<(v_{2m+1}+1)^2 \Rightarrow 0\leq v_{2m+1}\leq \sqrt{u_{2m+1}}<v_{2m+1}+1$$ or $$\frac{v_{2m}}{2m}\leq\frac{m\sqrt{2}}{2m}<\frac{v_{2m}+1}{2m} \Rightarrow 0\leq \color{blue}{\frac{\sqrt{2}}{2}}-\color{red}{\frac{v_{2m}}{2m}} <\frac{1}{2m}$$ $$\frac{v_{2m+1}}{2m+1}\leq \frac{\sqrt{2m^2+2m}}{2m+1}<\frac{v_{2m+1}+1}{2m+1} \Rightarrow 0\leq \color{blue}{\frac{\sqrt{2}}{2}}\cdot\frac{\sqrt{1+\frac{1}{m}}}{1+\frac{1}{2m}}-\color{red}{\frac{v_{2m+1}}{2m+1}}<\frac{1}{2m+1}$$ altogether (original sequence was partitioned into 2 subsequences, both converging) and by the squeeze theorem $$\lim\limits_{n\rightarrow\infty}\frac{v_n}{n}=\frac{\sqrt{2}}{2}=0.70710...$$

NOTE: I used the fact that if $n$ is the greatest positive integer such that $n^2 \leq m$, this also means $n^2\leq m < (n+1)^2$. If this is not true, then $(n+1)^2 \leq m$ and $n$ is not the greatest.