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.


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.


2 Answers 2


We simply have that


  • $\begingroup$ Is this the case where adding a finite number of terms won't affect the convergence of series ? Would that be a proper justification for concluding that the desired series converges too ? $\endgroup$
    – powerline
    Commented Aug 27, 2020 at 1:31
  • 1
    $\begingroup$ Yes exactly it is the same series without the first term. We could apply direct comparison test if we want but it is not necessary. The argument is $$\sum_1^\infty a_n=a_1+ \sum_1^\infty a_{n+1}=a_1+L\in \mathbb R$$ $\endgroup$
    – user
    Commented Aug 27, 2020 at 5:08

Let, for $ n\ge 1 $, $$S_n=\sum_{k=1}^na_k$$ $$=a_1+a_2+...+a_n$$ and $$T_n=\sum_{k=1}^na_{k+1}$$ $$=a_2+a_3+...+a_{n+1}$$ $$=S_n+a_{n+1}-a_1$$ then

$$\sum_{n\ge 1}a_{n+1} \;\text{converges} \; \implies \lim_{n\to +\infty}T_n\in \Bbb R \; \text{and} \; \lim_{n\to \infty}a_{n+1}=0$$

$$\implies \lim_{n\to\infty} S_n =\lim_{n\to\infty}T_n+a_1$$ $$\implies \sum_{n\ge 1} a_n \; \text{converges}$$

  • $\begingroup$ Shouldn't it be $lim_{n \to \infty}S_n=\lim_{n \to \infty}T_n - \lim_{n \to \infty}a_{n+1} - a_1$, and then since $a_{n+1}$ is decreasing and bounded, there is some $x\in \mathbb{R} $ such that $x=\lim_{n \to \infty}a_{n+1}$, and therefore $\sum_{n=1}^{\infty}a_n$ converges ? $\endgroup$
    – powerline
    Commented Aug 26, 2020 at 23:03
  • $\begingroup$ @GrigoriPerelman No, it is $ +a_1$. $\endgroup$ Commented Aug 26, 2020 at 23:17
  • $\begingroup$ @GrigoriPerelman $S_n=T_n-a_{n+1}+a_1$ $\endgroup$ Commented Aug 26, 2020 at 23:22
  • $\begingroup$ Yeah I made a mistake there its $+a_1$. But then $\lim_{n \to \infty}S_n=\lim_{n \to \infty}T_n -a_{n+1}+a_1$ and not $\lim_{n \to \infty}S_n=\lim_{n \to \infty}T_n + a_1$ as you said ? I mean $-a_{n+1}$ is missing in your post at that part, and then we have to prove the convergence for the sequence $a_{n+1}$ like I explained in first comment ? $\endgroup$
    – powerline
    Commented Aug 26, 2020 at 23:25

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