User:FUNG Cheok Yin/proof(i) A346526

Limit of the ratio of a(n) to a(n-1) in A346526

Statement: $\lim _{n\to \infty }{\frac {a_{n}}{a_{n-1}}}=1$, where a_n = A346526(n) (proposed by Stefano Spezia, Jul 22 2021).

Proof

The sequence is infinite; obviously we have $\lim _{n\to \infty }a_{n}=+\infty$ and $\lim _{n\to \infty }{\sqrt {a_{n}}}=+\infty$ .

Since $(10k)^{2}$, $(10k+1)^{2}$, $(10k+5)^{2}$ and $(10k+6)^{2}$ are definite terms in the sequence for non-negative integer $k$; therefore for arbitrary $n$, we have

$\lfloor {\sqrt {a_{n-1}}}\rfloor ^{2}<a_{n}\leq (\lfloor {\sqrt {a_{n-1}}}\rfloor +h)^{2}$,

where integer $h$ can be at least one member of {1, 2, 3, 4}.

Hence we can have $a_{n}=({\sqrt {a_{n-1}}}+g)^{2}$, where real number $g$ satisfies $0<g<5$.

(For example:

for $a_{n-1}=1925,{\sqrt {1925}}\approx 43.8748,\lfloor {\sqrt {a_{n-1}}}\rfloor =43$, we have $a_{n}=1956\approx (43+1.2267)^{2}\approx ({\sqrt {1925}}+0.3519)^{2}$;

for $a_{n-1}=37425,{\sqrt {37425}}\approx 193.4554,\lfloor {\sqrt {a_{n-1}}}\rfloor =193$, we have $a_{n}=37471\approx (193+0.5743)^{2}\approx ({\sqrt {37425}}+0.1189)^{2}$;

for $a_{n-1}=90000=300^{2}$, we have $a_{n}=90016\approx (300+0.0266)^{2}$;

for $a_{n-1}=92961,{\sqrt {92961}}\approx 304.89506,\lfloor {\sqrt {a_{n-1}}}\rfloor =304$, we have $a_{n}=92975\approx (304+0.9180)^{2}\approx ({\sqrt {92961}}+0.0230)^{2}$.)

$\lim _{n\to \infty }{\frac {a_{n}}{a_{n-1}}}$

= $\lim _{n\to \infty }{\frac {({\sqrt {a_{n-1}}}+g)^{2}}{a_{n-1}}}$

= $\lim _{n\to \infty }(1+{\frac {2g{\sqrt {a_{n-1}}}}{a_{n-1}}}+{\frac {g^{2}}{a_{n-1}}})$

= $\lim _{n\to \infty }(1+{\frac {2g}{\sqrt {a_{n-1}}}}+{\frac {g^{2}}{a_{n-1}}})$

= 1 + 0 + 0

= 1 .

$\square$

Written on Aug 12 2021