Theorem: If $\sum u_n$ and $\sum v_n$ are two positive terms series such that $\dfrac {u_n}{u_{n+1}} \geq \dfrac {v_n}{v_{n+1}} ~\forall n \geq m $, then, this is the same as $\sum u_n \leq \dfrac {u_m}{v_m} \sum v_n~~\forall~~n \geq m.$ Hence, $\sum u_n$ converges whenever $\sum v_n$ converges and $\sum v_n$ diverges whenever $\sum u_n$ diverges.
The Image below which show the proof for the test when $\lim\limits_{n \rightarrow \infty} n ~\left(\dfrac {u_n}{u_{n+1}} - 1 \right) >1$.
The next image shows the proof for the case when $p <1$.
The proof of Raabe's Test works on the basis of comparison with terms of the series $\sum\limits_{n=1}^{\infty} \dfrac {1}{n^p}. \sum\limits_{n=1}^{\infty} \dfrac {1}{n^p}$ diverges whenever $p \le 1$.
Then why is the Raabe's Test inconclusive for $p=1$? It should be divergent even then?
Thank you very much for your help in this regard.
$\endgroup$2 Answers
$\begingroup$Just browsing around questions related to Raabe's test. The questioner wasn't happy that Rabbe's test didn't do a better job of detecting divergence (although I guess he was happy enough with the convergence part of the test, weak though it is).
Let's rewrite Raabe's test in the way it should be presented.
Raabe convergence: Let $\{u_n\}$ be a sequence of positive numbers and suppose that $$\liminf\limits_{n \rightarrow \infty} n \left(\dfrac {u_n}{u_{n+1}} - 1 \right) > 1.$$ Then $\sum_{n=1}^\infty u_n$ converges.
Raabe divergence: Let $\{u_n\}$ be a sequence of positive numbers and suppose that $$n \left(\dfrac {u_n}{u_{n+1}} - 1 \right) \leq 1$$ for all sufficiently large $n$. Then $\sum_{n=1}^\infty u_n$ diverges.
In particular if $$\limsup\limits_{n \rightarrow \infty} n \left(\dfrac {u_n}{u_{n+1}} - 1 \right) < 1$$ then this divergence condition obviously holds. Why would one want to compute a limit instead of just using the inequality? Well sometimes limits are easier to compute than finding an inequality argument. That is the same answer for the ratio test after all. That test is just a comparison between a series and geometric series. If you can't handle the inequalities then the ratio test does it for you. Same here for the limit version of the Raabe test.
Should it happen that $$\limsup\limits_{n \rightarrow \infty} n \left(\dfrac {u_n}{u_{n+1}} - 1 \right) = 1$$ then you well know that the inequality could fail for infinitely many $n$ and so neither argument for the convergence or divergence case is available. The examples provided by Jack in the accepted answer show that one can't do any better if you feel somehow committed to Raabe. Raabe, who died in 1859 did some other important work at the time, but his name seems to survive mostly for this test. Most of us will be remembered for nothing. I hope he is happy with this small bit of fame.
$\endgroup$ $\begingroup$We have two sequences that satisfy: $$\lim_{n\to +\infty}n\left(\frac{u_n}{u_{n+1}}-1\right)=1,$$ namely $u_n = \frac{1}{n}$ and $u_n = \frac{1}{n\log^2 n}$, but while the series associated with the first one is divergent, the latter is convergent by Cauchy's condensation test, hence Raabe's test in inconclusive when the limit is one.
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