# Correct method of Proving Raabe's test?

I was wondering if my method of proof for Raabe's test was valid, since it is different from the normal method used with comparing to a sequence $\frac{1}{n^{p}}$ for some p > 1.

Raabe's Test (As described in Richard Beals' book Analysis An Introduction): Let $a_{n}$ be a positive sequence. Then $\sum_{n=1}^{\infty} a_n$ converges If $\lim_{n\rightarrow \infty }{\inf n(\frac{a_n}{a_{n+1}}-1)}>1$ and diverges if $n(\frac{a_n}{a_{n+1}}-1)\leq{1}$ $\forall n>N$

I said the following: Suppose $\lim_{n\rightarrow \infty }{\inf n(\frac{a_n}{a_{n+1}}-1)}>1$. Then $\exists{N}$ s.t $\forall{n}>N$ $\frac{n*a_n}{a_{n+1}}-n > 1$

Thus we have $\frac{a_{n+1}}{a_n} < \frac{n}{n+1}$. By taking the limit of both sides we get that by the ratio test, $\sum_{n=1}^{\infty} a_n$ converges. The second part is similar. My guess is that I cannot say for certain that $\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}$ exists, but I am unsure if this is the issue.

• You cannot apply the ratio test. For this, you need $\limsup_n a_{n+1}/a_n <1$. – PhoemueX Jan 22 '16 at 7:37
• – B. S. Thomson Jan 22 '16 at 16:42

The proof for divergence goes as follows:

We choose a positive number $\epsilon$ such that $l+\epsilon<1$.

Since $\lim_\limits{n\to\infty}n\left(\frac{u_n}{u_{n+1}}-1\right)=l$, then for any $\epsilon>0$ , there exists a natural number $m$ such that $$\left| \,\ n\left(\frac{u_n}{u_{n+1}}-1\right)-l\,\ \right|< \epsilon \,\ \forall \,\ n\ge m$$

Therefore, we have
$$l-\epsilon<n\left(\frac{u_n}{u_{n+1}}-1\right) <l+\epsilon \,\ \forall \,\ n\ge m$$ Now say $$l+\epsilon=r$$ Hence $$r<1$$ So we can say that $$n\left(\frac{u_n}{u_{n+1}}-1\right) <r$$ or, $$\frac{u_n}{u_{n+1}} < \frac{r}{n}+1$$

Rewrite this as $$n \frac{u_n}{u_{n+1}} -(n+1) < r-1 \leq 0.$$ Then observe that $$nu_n \leq (n+1)u_{n+1}$$ for all $n$ greater than $m$. Thus we have an increasing sequence $\{nu_n\}$ and, in particular there is some positive constant $c$ for which $nu_n\geq c$.

Thus for $n\geq m$ we have $$u_n \geq \frac{c}{n}$$ and a comparison with the harmonic series establishes divergence as you desired.

There is a more general version of Rabbe's test known as Kummer's test with very much the same proof. For that you assume there is a sequence of positive numbers $D_n$ and you compute the limit $$L =\lim_{n\to\infty} \left[ D_n\frac{u_n}{u_{n+1}} -D_{n+1} \right]$$

Raabe's test is just Kummer's test with $D_n=n$.

The divergence part of Kummer's test (and hence also Raabe's test) doesn't actually require limits. You simply need that $$D_n\frac{u_n}{u_{n+1}} -D_{n+1} \leq 0$$ for all sufficiently large $n$ and the divergence of $\sum_{n=1}^\infty 1/D_n$ and you can conclude divergence.