# Proof of Raabe's test

Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$ Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?

P.S. using Kummer's test doesn't count

If $L>1$, choose $\epsilon\gt 0$, $L-\epsilon>1$ then

$$1-\frac{L-\epsilon}{n} > \frac{a_{n+1}}{a_n}$$

Choose $p$ such that $1\lt p\lt L-\epsilon$, $\sum\frac1{n^p}$ converges. $b_n=\frac1{n^p}$, if $n$ big enough, then

$$\frac{b_{n+1}}{b_n}=(1-\frac{1}{n+1})^p=1-\frac{p}{n}+O(\frac1{n^2})\gt 1-\frac{L-\epsilon}{n}> \frac{a_{n+1}}{a_n}$$

so $\sum a_n$ converges

• How is $(1-\frac{1}{n+1})^p=1-\frac{p}{n}+O(\frac1{n^2})$? Shouldn't it be $p/(n+1)$? Commented Nov 20, 2023 at 18:37

The series even converges absoute, thus set $a_n:=\left|a_n\right|$

Now (if $n$ is big enough) $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right) \Longleftrightarrow \frac{a_{n+1}}{a_n}\leq\frac{n-L}{n}$$

This is equivalent to $$\left(L-1\right)a_n \leq \left(n-1\right)a_n-na_{n+1}$$

Because of $L>1$ the left side is bigger than zero, so

$$0\leq \left(n-1\right)a_n-na_{n+1}\Longleftrightarrow na_{n+1}\leq \left(n-1\right)a_n$$

That means that $a_n$ is decreasing and bounded, so it does converges. Now define the sum

$$\sum b_n = \sum \left(n-1\right)a_n-na_{n+1}$$

Which is a telescop-sum and thus it converges

But that implies that the sum over $a_n$ $$\sum a_n \leq \sum \left(L-1\right)a_n \leq \sum b_n$$ is convergent as well by the comparison test.

The following is a new argument for the convergence part. Assume that $$L>1$$. Choose $$\varepsilon > 0$$ small enough such that $$L-\varepsilon > 1.$$ There exists some $$1 \ll N=N(\varepsilon)$$ such that $$n\Big( 1- \frac {a_{n+1}}{a_n} \Big) > L-\varepsilon$$ for any $$n \geq N$$, namely $$\frac {a_{n+1}}{a_n} < 1 - \frac{L-\varepsilon}n = \frac{n-(L-\varepsilon)}n$$ for any $$n \geq N$$. Since $$L-\varepsilon>1$$, one can always choose $$\alpha >1$$ in such a way that $$\frac{n-(L-\varepsilon)}n < \Big( \frac{n-1}n \Big)^\alpha$$ for any $$n \gg 1$$, say $$n \geq M > N$$. Hence for large $$n$$, we obtain $$a_{n+1} \leq \Big( \frac{n-1}n \Big)^\alpha a_n \leq \Big( \frac{n-2}n \Big)^\alpha a_{n-1} \leq \cdots \leq \Big( \frac{M-1}n \Big)^\alpha a_M.$$ Hence $$\sum a_n$$ converges by the comparison test.