This will most likely be on the exam, but it is not given in the text book. In my notebook I have this proof which I will type out, but it makes no sense. Here it goes:
$$\text{D'Alembert Criterion}$$
Let $\sum_{n=1}^{\infty}a_n$ with $a_n > 0$.
1.) If $\exists N$ such that $\forall n >N$, $\frac{a_{n+1}}{a_n}\leq q < 1$ then $\sum_{n=1}^{\infty}a_n$ converges.
2.) If $\exists N$ such that $\forall n >N$, $\frac{a_{n+1}}{a_n}> 1$ then $\sum_{n=1}^{\infty}a_n$ diverges.
Proof:
$$ \forall n >N \ \ \ \frac{a_{n+1}}{a_n}\leq q < 1 \\
For \ \ \ n=1: \frac{a_{2}}{a_1} \leq q \implies a_2 \leq qa_1
\\
For \ \ \ n=2: \frac{a_{3}}{a_2} \leq q \implies a_3 \leq q^2a_1
\\
For \ \ \ n=2: \frac{a_{3}}{a_2} \leq q \implies a_4 \leq q^3a_1$$
$$\\ …. \\
\\
For \ \ \ \frac{a_{n+1}}{a_{n}} \leq q \implies a_3 \leq q^{n-1}a_1=\frac{a_1}{q}q^n
$$
So if $\sum_{n=1}^{\infty}q^n = \frac{q}{1-q}$ for $|q|<1 \implies a_n conv.???$
Best Answer
The idea used in the attempted proof of (1) is along the right lines, but there are problems in the details. First, to get in the mood, let us knock off the easier (2).
We suppose that there is an $N$ such that if $n\ge N$ then $\frac{a_{n+1}}{a_n}\ge 1$. Then in particular $a_{N+1}\ge a_N$, and $a_{N+2}\ge a_{N+1}\ge a_N$, $a_{N+3}\ge a_{N+2}\ge a_N$, and in general if $n\gt N$ then $a_n\ge a_N$.
It follows that the terms $a_n$ cannot have limit $0$. Thus, by the simplest divergence test, $\sum_1^\infty a_n$ diverges.
We now attack (1), first giving a restated version of it. We will suppose thar there exists a $q\lt 1$, and an integer $N$, such that if $n\ge N$ then $\frac{a_{n+1}}{a_n}\le q$. Consider the partial sums $S_k=\sum_1^k a_n$. The sequence $(S_k)$ is increasing. If we can show that it is bounded above, then the sequence $(S_k)$ of partial sums converges, which is exactly what it means for $\sum_1^\infty a_n$ to converge.
Note that as in your calculation we have $a_{N+1}\le a_Nq$, and $a_{N+2}\le a_Nq^2$, and $a_{N+3}\le a_Nq^3$. In general, if $k\ge N$, then $a_k \le a_Nq^{k-N}$. We conclude that if $k\ge N$ then $$S_k\le a_1+a_2+\cdots +a_{N-1}+\sum_{i=0}^{k-N} a_N q^i.$$ Since $q\lt 1$, the series $\sum_{i=1}^\infty a_N q^{i-N}$ is an infinite geometric series with common ratio $q\lt 1$. It follows from Inequality (1) that if $k\ge N$ then $$S_k \le a_1+\cdots +a_{N-1}+\frac{a_N}{1-q},$$ which shows that the sequence $(S_k)$ of partial sums is bounded. This completes the proof.
Remark: We have been semi-formal, and that may lead one to lose contact with the informal reason why $\sum a_n$ converges. If there is a $q\lt 1$ such that after a while $\frac{a_{n+1}}{a_n}\le q$, then after a while the terms of our series are $\le$ to the terms of a convergent geometric series, so by Comparison $\sum_1^\infty a_n$ converges.