# State and prove Raabe's Test.

## Expert Answers

Raabe's test is the ratio test for convergence of a series.

Consider the limit `lim_(n->) |a_(n+1)/a_n| = L`

Raabe's test says that if `L<1` then the series converges. If `L>1` then the series diverges. If `L=1` the test is inconclusive.

Proof:

We proceed by applying the limit comparsion test. This says ` `that if the limit`lim_(n->oo) |a_n|/|b_n|`  exists and is non-zero, then `Sigma_1^oo |a_n|` converges if and only if `Sigma_1^oo |b_n|` converges.

Note that if `|a_n|` converges then `a_n` converges.

We compare `a_n` to the geomertic series `r^m` where

`r = (L+1)/2`

Now, if `L<1` then `r>L` and `|a(n+1)| < r|a_n|`  for `n>N` (`n` large enough)

By induction

`|a_(n+m)| <r^m|a_n|` for `n>N` and `i>0`

`implies`  `Sigma_(N+1)^oo |a_m| = Sigma_1^oo |a_(N+m)| < Sigma_1^oo r^m|a_N|`

`implies Sigma_1^oo |a_(N+m)| < |a_N| Sigma_1^oo r^m = |a_N|(r/(1-r)) < oo`

because `r<1`

Therefore if `L<1` the series `a_n` converges absolutely (`Sigma_1^oo a_n ` converges if `Sigma_1^oo |a_(N+m)|` converges) .

If `L>1` t` `hen `|a_(n+1)| > |a_n|`  for ` ``n>N` . Such a series cannot converge, so the series diverges when `L>1`.

If `L=1` we cannot show whether `lim_(n->oo) |a_n|/|b_n|` (where `b_n` is a geometric series) exists and is non-zero or not, so the test is inconclusive.

That concludes the proof.

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