Proof of Raabe's test

Solution 1:

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

Solution 2:

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.

Solution 3:

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.