I do not understand an inequliaty that is used to prove the ratio test

Question

I do understand:

$\exists q\lt1:\forall k\in\Bbb N_{0}:\lvert\frac{a_{k+1}}{a_{k}}\rvert \leq q$ is equivalent to $\lvert a_{k+1}\rvert \leq q \cdot \lvert a_{k}\rvert$

Now the statement is that $ \sum \limits_{k=0}^{\infty}a_{k}$ converges absolutely.

The proof is what I do not understand:

"Inductively you can show: $\forall k \in\Bbb N_{0}: \lvert a_{k}\rvert \leq q^{k}\cdot\lvert a_{0}\rvert$ "

How do we go from $$\lvert a_{k+1}\rvert \leq q \cdot \lvert a_{k}\rvert$$ to $$\lvert a_{k}\rvert \leq q^{k}\cdot\lvert a_{0}\rvert$$

And how did $q^{k}$ happen?

Answer 1

"Inductively"

You have proven that for ever $k$ that $|a_{k+1}| \le q\cdot|a_{k}|$.

So that would mean $|a_1| \le q\cdot |a_0|$. But $|a_2| \le q \cdot|a_1|$.

And $|a_1| \le q\cdot |a_0|$ so $|a_2| \le q\cdot|a_1| \le q\cdot (q\cdot |a_1|) = q^2\cdot |a_0|$

And $|a_3| \le q\cdot |a_2| \le q\cdot (q^2|a_0|)=q^3\cdot |a_0|$.

And so on.

If you have $|a_{k}| \le q^{k}|a_0|$ then $|a_{k+1}| \le q\cdot |a_k|\le q(q^k\cdot |a_0|)= q^{k+1}|a_0|$.

That's a proof by induction.

Answer 2

Note that $|a_{k+1}|\leq |a_k|q$, and as this is true for all $k$, you change $k$ to $k-1$ and get $|a_k|\leq |a_{k-1}|q$, and merging it with the previous inequality, we have $|a_{k+1}|\leq q^2|a_{k-1}|$.

Similarly, you replace $k$ with $k-2,k-3,...,0$. Can you get the idea from here?

Answer 3

Heuristically speaking $|a_0| \geq q|a_1| \geq q(q|a_2|) = q^2|a_2| \geq q^2(q|a_3|) = q^3|a_3|$ and so on until you reach your desired $k$.

More carefully you can use the principle of induction; to prove $|a_0| \geq q^k|a_k|$ for all $k$ it suffices show that $|a_0| \geq q|a_1|$ and that if $|a_0| \geq q^k|a_k|$ for some $k$ then $|a_0| \geq q^{k+1}|a_{k+1}|$.