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.*

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**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.