How can I prove this sequence is decreasing WITHOUT using induction?

Question

I am given a sequence $a_1=\ln(5)$ and $a_n=\ln\left( \frac{e^{2a_{n-1}}+16}{10}\right)$, for $n\geq 2$ and I am asked to show that $a_n \geq 0$ for all $n \geq 1$.

I can do this easily with mathematical induction.

Base case: $\ln(5) \geq0$

Induction hypothesis: $a_k \geq 0$ for all $n=k$

Induction step:

$a_k \geq 0$

$2a_k \geq 0$

$e^{2a_k} \geq 1$

$e^{2a_k} + 16 \geq 17$

$\frac{e^{2a_k} + 16}{10} \geq \frac{17}{10}$

$\ln\left( \frac{e^{2_{a_k}}+16}{10}\right) \geq \ln \left( \frac{17}{10} \right) \geq 0$

$a_{k+1} \geq 0$

So because I have shown that my induction hypothesis is true for all $n=k+1$, I'm done.

HOWEVER, this is supposedly possible to do WITHOUT mathematical induction. How would I go about doing that? For context, this is an intro level calculus course that has not covered the concept of a limit yet.

Answer 1

$e^{x} >0$ for all real numbers $x$ so $a_n > ln(\frac {16} {10}) >ln 1=0$.

Answer 2

We have $a_1 = \ln(5) \geq 0$, so we only need to show $a_n \geq 0$ for all positive integers $n \geq 2$.

Notice that the function $f(x) = \ln x$ is an increasing function on its domain. Furthermore, $e^x \geq 0$ for any $x \in \mathbb{R}$. Therefore, for any positive integer $n > 1$, $$\ln \left(\frac{e^{2a_{n-1}} + 16}{10}\right) \geq \ln \left(\frac{0 + 16}{10}\right) = \ln(1.6) > \ln(1) = 0.$$

So $a_n \geq 0$ for all positive integers $n$.