0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.