Dgdrxrt

Please give me a hint for limit of $lim_ntoinfty3^2^n$ in $mathbb Q_5$.

First, new absolute value $|cdot|'$ on $mathbb Q$ is defined as the following:

For $fracnmin mathbb Q$,$$|fracnm|'=5^-v$$
if $fracnm=5^vp_1^v_1p_2^v_2p_3^v_3...p_r^v_r$ where $p_i$: distinct prime number without 5 and $v_iinmathbb Z$.

Then, the absolute value satisfies $|r|'=0
Leftrightarrow r=0$ and triangle inequiality. Secondly, based on this absolute value and equivalence relation for Cauchy sequence, new $mathbb Q_5$
system is constructed.

This is a moderately delicate question. Let’s look at $3^2^n+2$ modulo $25$. We have:
$$
3^2^n+2=3^4cdot2^n=81^2^n=(1+5cdot16)^2^nequiv(1+5)^2^nequiv1+2^ncdot5pmod25
$$
Now, you can check that for $nequiv0,1,2,3pmod4$, you get $2^nequiv1,2,4,3pmod5$ respectively, in other words, $3^2^n+2equiv6,11,21,16pmod25$ respectively, so there’s nothing like a limit here.

In fact the situation is even worse (or better, depending on your taste). Using the $5$-adic logarithm, you can show that when you restrict to values of $nequiv0pmod4$, the numbers $(1+5cdot16)^2^n$ are dense in $6+25Bbb Z_5$, and similarly for $nequiv1pmod4$, the numbers $(1+5cdot16)^2^n$ are dense in $11+25Bbb Z_5$, etc. But this density question involves more work than you want to see now.

The moral is: no limit.

By repeated squaring we have $3^4equiv 6bmod 25$, $3^8equiv 11$, $3^16equiv 21$, $3^32equiv 16$, $3^64equiv 6$ so the expression cycles $bmod 25$. The terms in the cycle differ by amounts having $5$-adic norm $1/5$ which never goes to zero. So there cannot be a $5$-adic limit.

Since you're repeatedly squaring in the limit, if the limit exists,

$$lim_nto infty 3^2^n = L$$

then it must end up satisfying $L^2=L$. So either L=0 or L=1. It's not 0 because 3 is not divisible by 5, so it must be 1 if it exists. That means this limit must be zero:

$$lim_n to infty 3^2^n - 1 = 0$$

We can factor that pretty easily by a geometric series, for reference:

$$frac3^2^n-13-1 = sum_k=0^2^n-1 3^k$$

so we can rewrite the previous limit as,

$$lim_n to infty 2*sum_k=0^2^n-1 3^k = 0$$

in the limit if this sum converges (it doesn't), it should converge to:

$$2 * frac11-3$$

which is not 0, so the limit does not exist.