Ratio test (missing proof step): $\lim_\limits{n\rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|<1 \Rightarrow \sum_\limits{k=0}^n a_k$ converges

Question

I'm trying to proof the ratio test for convergence of sum of a sequence:

$l :=\lim_\limits{n\rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right| <1 \Rightarrow s_n := \sum_\limits{k=0}^n a_k$ converges absolutely

Currently I'm stuck proving it, and online I've seen other proofs, but I'm interested in this one explicitly, since it's the most intuitive for me:

I'm using the absolute value as a norm for the limit definition.

$\lim_\limits{n\rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|<1 \Leftrightarrow \forall\epsilon>0 \quad \exists n_0>0: \left|\left|\frac{a_{n+1}}{a_n} \right| - l\right| < \epsilon \quad \forall n\geq n_0$

So we have:

$\forall\epsilon>0 \quad \exists n_0>0: \left|\left|\frac{a_{n+1}}{a_n} \right| - l\right| < \epsilon \quad \forall n\geq n_0$

MISSING STEP / ASSUMPTION

$\forall\epsilon>0 \quad \exists n_0>0:\left|\frac{a_{n+1}}{a_n} \right| - l < \epsilon\quad \forall n\geq n_0$

$\forall\epsilon>0 \quad \exists n_0>0:\left|\frac{a_{n+1}}{a_n} \right| < \epsilon+ l\quad \forall n\geq n_0$

we choose $\epsilon : \epsilon + l < 1$ with its corresponding $n_0$

$\left|\frac{a_{n+1}}{a_n} \right| < \epsilon + l \quad \forall n>n_0$

And we write:

$\left|a_n\right| = \left|\frac{a_{n}}{a_{n-1}}\right|\left|\frac{a_{n-1}}{a_{n-2}}\right|...\left|\frac{a_{n_0+1}}{a_{n_0}}\right|\left|a_{n_0}\right| < (e+l)^{n-n_0+1} \left|a_{n_0}\right|$

Therefore:

$ \lim_\limits{n\rightarrow \infty} s_n= \sum_\limits{k=0}^\infty \left|a_k\right| = \sum_\limits{k=0}^{n_0-1} \left|a_k\right| + \sum_\limits{k=n_0}^{\infty} \left|a_k\right| < \sum_\limits{k=0}^{n_0-1} \left|a_k\right| + \sum_\limits{k=n_0}^{\infty}(e+l)^{k-n_0+1} \left|a_{n_0}\right| = C < \infty \quad \blacksquare$

I don't understand why I can take the assumption mentioned above? I feel like there's an obvious link I'm missing.

Answer 1

The question was why

$\forall\epsilon>0 \quad \exists n_0>0: \left|\left|\frac{a_{n+1}}{a_n} \right| - l\right| < \epsilon \quad \forall n\geq n_0$

would imply the following:

$\forall\epsilon>0 \quad \exists n_0>0:\left|\frac{a_{n+1}}{a_n} \right| < \epsilon+ l\quad \forall n\geq n_0$

The key is the rule $|x|<a\Leftrightarrow -a<x<a$ for $a>0$. If we apply the rule to $\left|\left|\frac{a_{n+1}}{a_n} \right| - l\right| < \epsilon$, we have

$\displaystyle -\epsilon < \left|\frac{a_{n+1}}{a_n} \right| - l < \epsilon\tag*{}$

This implies

$\displaystyle \left|\frac{a_{n+1}}{a_n} \right| - l<\epsilon \tag*{}$

or

$\displaystyle \left|\frac{a_{n+1}}{a_n} \right| <\epsilon+l \tag*{}$

Answer 2

You are assuming that$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=l<1.$$Take $\varepsilon>0$ such that $l+\varepsilon<1$. Since $\varepsilon>0$,$$(\exists n_0\in\Bbb N)(\forall n\in\Bbb N):n\geqslant n_0\implies\left|\left|\frac{a_{n+1}}{a_n}\right|-l\right|<\varepsilon.$$ But then, $$n\geqslant n_0\implies\left|\frac{a_{n+1}}{a_n}\right|<l+\varepsilon.$$So, if $n\geqslant n_0$,$$|a_n|\leqslant\left|a_{n_0}\right|(l+\varepsilon)^{n-n_0}.$$