Is it possible to find $\limsup\limits_{n\to\infty} \frac{2×3^n-3×2^n}{2^{\alpha(n)}-3^n}$?

Question

I need help to solve this problem:

$$\lim_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n}{2^{\alpha(n)}-3^n}$$ or

$$\limsup_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n } {2^{\alpha(n)}-3^n}$$

where, $\alpha(n)-$ is the smallest positive integer, such that $2^{\alpha(n)}-3^n>0$.

I tried $\alpha (n)= \lfloor n\log_23+1 \rfloor$. Then, we have

$$\lim\limits_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n}{2^{\alpha(n)}-3^n}= \lim\limits_{n\to\infty} \frac{2}{\frac{2^{\lfloor n\log_23+1 \rfloor}}{3^n}-1}$$ I'm stuck here. I can not continue..Can MSE help me? Are my steps wrong? Can I take $\alpha (n)= \lfloor n\log_23+1 \rfloor$ ?

Thank you.

Answer 1

Just some numerical results.

I wrote this simple-minded python script

from math import log, ceil

log2=log(2)
log3=log(3)
alpha = lambda n:ceil(n*log3/log2)

def f(n):
    return (2*3**n - 3*2**n)/(2**alpha(n)-3**n)

for n in range(100,200):
    print(n, f(n))

which produces the output

100 4.785899883385597
101 2.245864550967373
102 7.681953701600157
103 2.9391568861026434
104 16.62351490582627
105 4.050643858783844
106 2.016810341215728
107 6.101716995999464
108 2.5961591759629563
109 11.095955317058582
110 3.486378701508524
111 40.72703423804435
112 5.014920551792514
113 2.311925239267859
114 8.217379797557589
115 3.0402123771418967
116 19.005531878533308
117 4.222586386526414
118 2.0728442833578278
119 6.45358628404818
120 2.6790249107257487
121 12.168397114275404
122 3.6200389288236945
123 57.16436556434427
124 5.263401099906297
125 2.381018741841479
126 8.824137017681315
127 3.146956188626232
128 22.134545588402613
129 4.407143726153516
130 2.1312610654441864
131 6.8428854112276705
132 2.7661274293951537
133 13.45095512434899
134 3.7623374007474766
135 94.98349540686164
136 5.533905896179148
137 2.4533520206054322
138 9.51742886535484
139 3.259871847121801
140 26.426816770405463
141 4.605738069827515
142 2.1922102669346684
143 7.275880242612633
144 2.8577915486596823
145 15.011926476150077
146 3.9141250929274456
147 273.476284960899
148 5.82947053601099
149 2.529151287823254
150 10.317109843664202
151 3.3794992545397133
152 32.678438624915465
153 4.820013706951867
154 2.2558542371938963
155 7.7603178081785575
156 2.954376137320582
157 16.95280440377464
158 4.076368522620288
159 2.0252951944873963
160 6.153715296079386
161 2.6086642979673247
162 11.249611265589706
163 3.5064431516181305
164 42.62661733634982
165 5.051882141737799
166 2.3223694882668457
167 8.305891453539244
168 3.0562786992193884
169 19.431236934196654
170 4.250170048286752
171 2.081687045094281
172 6.5109930257336694
173 2.6921629750026606
174 12.350910351424579
175 3.641383153327096
176 60.9237940790063
177 5.303578674949636
178 2.3919482740096836
179 8.924893031795749
180 3.1639407167994875
181 22.70609833805777
182 4.436792604793172
183 2.1404840037695347
184 6.906584320141273
185 2.7799464337954323
186 13.671267446609711
187 3.7850857076885367
188 105.67242411788106
189 5.577733998614978
190 2.464800385329663
191 9.63314618545789
192 3.2778539067755394
193 27.234272434871727
194 4.637690047702137
195 2.201837533693299
196 7.346958603409351
197 2.872344466392401
198 15.283095133865157
199 3.938418422291965

I've tried it for other ranges, up to about $n=5200,$ with similar results, so I'm willing to guess that the limit doesn't exits. As for the lim sup, I haven't a clue. The largest value I noticed was $f(5161)\approx 337.115.$

EDIT

On second thought, I think the lim sup is probably $\infty.$ You might take $\alpha(n)$ and $n$ to be the numerators and denominators, respectively, of a sequence of rationals that converges to $\log_2{3}.$ I thought about the convergents of the continued fraction, but I don't see that this buys you anything.

Answer 2

I don't think the limit exists. Write $3^n=2^{n\log_2(3)}$, then the latter simplifies to $$ \frac{2}{2^{\lfloor n\log_2(3) + 1 \rfloor - n\log_2(3)}-1} = \frac{2}{2^{\lfloor n\log_2(3) + 1 \rfloor - n\log_2(3)}-1} $$ I believe there exists an increasing sequence $(a_n)_{n=1}^{\infty}\in\mathbb{N}^{\mathbb{N}}$ such that $\lfloor a_n\log_2(3) + 1 \rfloor - a_n\log_2(3)\to 0$. The limit under this subsequence will thus converge to infinity. Moreover, we can also pick the $(a_n)$ such that $\lfloor a_n\log_2(3) + 1 \rfloor - a_n\log_2(3)\to1$, which means the subsequence converges to $1$. Thus, the limit can't exist.

I am $99\%$ sure that this is a valid argument. However, to be precise, you have to define the $a_n$ in both cases and work out their respective limits.

Answer 3

We have $$\alpha(n) = \lceil n \log_2 3 \rceil = n \log_2 3 + 1 - \{n \log_2 3\}, \\ \frac {2 \cdot 3^n - 3 \cdot 2^n} {2^{\alpha(n)} - 3^n} = \underbrace {3^{-n} (3 \cdot 2^n - 2 \cdot 3^n)}_{\to -2} \cdot \frac 1 {1 - 2^{1 - \{n \log_2 3 \}}} \,.$$ Since $\{n \beta\}$ is dense in $[0, 1]$ for irrational $\beta$, we can choose a subsequence that doesn't converge and can choose a subsequence that converges to $1$.