# What convergence criterion was used in this proof?

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.

We simply have that

$$\sum_{n=1}^{\infty}a_{n+1}=\sum_{n=2}^{\infty}a_{n}=-a_1+\sum_{n=1}^{\infty}a_{n}$$

• 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 ? Aug 27, 2020 at 1:31
• 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 $$\sum_1^\infty a_n=a_1+ \sum_1^\infty a_{n+1}=a_1+L\in \mathbb R$$
– user
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

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

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

• 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 ? Aug 26, 2020 at 23:03
• @GrigoriPerelman No, it is $+a_1$. Aug 26, 2020 at 23:17
• @GrigoriPerelman $S_n=T_n-a_{n+1}+a_1$ Aug 26, 2020 at 23:22
• 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 ? Aug 26, 2020 at 23:25