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I want to prove the ratio test

$$\lim \sup \left| \frac{a_{n+1}}{a_n} \right|<1$$

implies the Raabe's test

$$\lim \sup \left| \frac{a_{n}}{a_{n+1}} \right|>1 + \frac{C}{n}$$

for $C>1$

I am doing the following:

$$\lim \sup \left( \left| \frac{a_{n}}{a_{n+1}} \right|-1\right)n>C$$

$$ \left(\left| \frac{a_{n}}{a_{n+1}}\right|-1\right)n >p+\frac{p(p-1)}{2n} + \dots$$

$$\left| \frac{a_{n}}{a_{n+1}}\right| > 1 + \frac{p}{n}+\frac{p(p-1)}{2n^2} + \dots$$

$$\left| \frac{a_{n}}{a_{n+1}}\right| > \left( \frac{n+1}{n} \right)^p >1$$

since $\sum \frac{1}{n^p}$ converges if $p>1$ then by the ratio test it will converge if the fraction is $>1$. Therefore the Raabe's test gives convergence.

Is this correct?

user186918
  • 335

1 Answers1

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What you have given as Raabe's test is not correct. Raabe's direct test states that defining

$C_n = n(a_n/a_{n+1}-1)$

Then $\liminf C_n>1$ implies convergence and $\limsup C_n<1$ implies divergence.

Paul R.
  • 141