# Souvenirs from Bandung part 2 (and Bulgaria) :D

When I packed my bag to go to Bandung on Wednesday I only put in one shirt for Thursday although I planned to stay until Friday. I thought, “hey, it’s Bandung and I travel there to shop, why would I pack so many shirts? I should just go buy some”. I probably shouldn’t have oversimplified things. Though one of the reason I went to Bandung is to shop, my shopping time frame is pretty tight and sometimes I can be a bit picky. Thus, I fail to find even a single suitable shirt~ when the shirt match, the price doesn’t and vice versa 😆 Luckily I met my junior Oskar who gave me souvenirs from his short visit in Bulgaria for International Mathematics Competition. Why am I so lucky? Because one of the souvenir is T-shirt, yeaaay, problem solved 😉

Thank you so much Os 🙂 Here’s the photo of the souvenirs:

Souvenirs from Bulgaria is not the only thing he gave me; He also gave me a little math problem.

If you can’t see the photo clearly, the problem sounds like this:

Let $a_n$ be a converging sequence of positive real numbers that satisfy the following equation

$a_n^{a_n} = e^{a^{n+1}}$

Proves that $a_n$ is the constant sequence $e$. The problem is really intuitive isn’t it? The answer to this problem is also quite lucid. It took me less than a minute to solve. You may wanna try to tackle the problem yourself before you see my solution below.

Proof: If $a_n$ is the constant sequence $e$ then it clearly converges and satisfy the equation. Assume that $a_n$ is not the constant sequence $e$. Then there’s a member of the sequence that is not $e$. Without loss of generality, pick that as $a_1$ and also assume that $a_1 > e$. Why don’t we lose generality by stating both assumptions?
The equation above can be rewritten as $\ln a_n = \frac{a_{n+1}}{a_n}$ (why?). Since $a_1 > e$, we have $a_2 > a_1$. But then, $\ln a_2 > \ln a_1$.
By repeating this step we have $\ln a_{n} > ... > \ln a_2 > \ln a_1$.
Therefore, $\frac{a_{n+1}}{a_n} > ... >\frac{a_3}{a_2} > \frac{a_2}{a_1}$.
Write $a_{n+1} = a_1 (a_2/a_1) (a_3/a_2) ... (a_{n+1}/a_n)>a_1(a_2/a_1)^n$.
Since $\frac{a_2}{a_1} > 1$, we know $a_n$ diverge to infinity.

Q.E.D