I have this proof of Raabe criterion for convergence in my notebook, but only 1 final conclusion is what I fail to understand(since it doesn't mention what criterion was used) so I will keep it concise with the other parts.
Statement: If $\lim_{n \to \infty}n\cdot(\frac{a_n}{a_{n+1}}-1)=\alpha>1$, then the series $\sum_{n=1}^\infty a_n$, $a_n>0$, converges.
Proof.
We pick a $q \in \mathbb{R}$ such that $\alpha > q > 1 $. Eventually, there is some $K$ so that for all $n>K$ we get $$na_n-(n+1)a_{n+1} > (q-1)a_{n+1} > 0\tag{$*$}$$
hence $na_n > (n+1)a_{n+1}$. Therefore, the sequence $\{na_n\}_{n=1}^{\infty}$ is decreasing and bounded, so it means it converges. Now, let $\lim_{n \to \infty}na_n=c$.
From $(*)$ it follows $a_{n+1}<\frac{na_n-(n+1)a_{n+1}}{q-1}$. We observe the series $S=\frac{1}{q-1} \sum_{n=k}^{\infty}na_n-(n+1)a_{n+1}$. Eventually, the sequence of partial sums $S_n=ka_k-(k+n)a_{k+n}$, and $S=\lim_{n \to \infty}ka_k-(k+n)a_{k+n}=ka_k-c$, and so the series $S$ converge, and since $a_{n+1}<\frac{na_n-(n+1)a_{n+1}}{q-1}$, (by comparison criterion?) the series $\sum_{n=1}^{\infty}a_{n+1}$ converge, hence $\sum_{n=1}^{\infty}a_n$ converges.
End of proof.
My question:
How does the convergence of $\sum_{n=1}^{\infty}a_n$ follow from $\sum_{n=1}^{\infty}a_{n+1}$ ? We know $na_n>(n+1)a_{n+1}$ and by that $a_n>a_{n+1}$, so it shouldn't be the comparison criterion which enabled us to make the conclusion, right ?
Note: The proof is very messy written in the notebook, so there may be some mistakes, sorry for those.