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:

an AUBG shirt and a keychain

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

the 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.



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