Comparison of series test Is there a series that is conclusive by Ratio test, but not Raabe's test?
When the Ratio test was inconclusive, i used Raabe's test.
I wonder which of the test is more stronger.
Give some examples or comments! Thank you!
 A: 
Is there a series that is conclusive by Ratio test, but not Raabe's
  test?

If the ratio test is conclusive, then Raabe's test gives the same result.  
Recall that the basic ratio test applied to a series $\sum a_n$ where terms are not exclusively non-negative real numbers states that if
$$\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} = L,$$
then the series is convergent (and absolutely convergent) if $L < 1$ and divergent with $a_n \not\to 0$ if $L > 1$.
Raabe's test is also applicable if terms are not exclusively non-negative real numbers, where 
$$\lim_{n \to \infty}n \left(\frac{|a_n|}{|a_{n+1}|} - 1 \right) =  \lambda,$$
implies convergence for $\lambda > 1$ and divergence for $\lambda < 1$.  The conclusion also holds if we interpret $\lambda >(<)\,  1$ to include the case $\lambda = +\infty(-\infty)$. In this way Raabe's test leads to the same conclusion when the ratio test is conclusive.
For example, suppose the ratio test leads to the conclusion of convergence with $L <1$.  We then have
$$\lim_{n \to \infty} \frac{|a_n|}{|a_{n+1}|}- 1 = \frac{1}{L} - 1 > 0,$$
whence,
$$\tag{*}\lim_{n \to \infty} n\left(\frac{|a_n|}{|a_{n+1}|}- 1\right) = +\infty$$
To see that (*) implies (absolute) convergence, note that for any $R> 0$ there exists a positive integer $N$ such that for all $n \geqslant  N$ we have
$$n\left(\frac{|a_n|}{|a_{n+1}|}- 1\right) > 1 +R,$$
which implies
$$n |a_n| - (n+1) |a_{n+1}| > R|a_{n+1}|$$
Summing from $n = N$ to $M-1$ we get
$$\sum_{n=N+1}^M|a_n| = \sum_{n=N}^{M-1}|a_{n+1}| < R^{-1}(N|a_N| - M|a_M|) \leqslant R^{-1}N|a_N|.$$
This proves that the sum on the LHS is bounded and increasing for all $M$, and, hence, $\sum |a_n|$ is convergent (both by the ratio test and Raabe's test).
A: There are some key differences between both tests:


*

*Ratio test deal with real or complex series with non vanishing terms.

*Raabe-Duhamel's test with positive real series only.


Therefore, you may use ratio test in cases where Raabe test can't be used. One example is the series $\sum \frac{e^{in}}{2^n}$. Ratio test enables to conclude that the series converges while Raabe test can't be used.
Now, providing that you restrict your consideration to real series with positive terms $\sum a_n$:


*

*Ratio test is conclusive for $L \neq 1$ where $L = \lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}$

*Raabe test provides a conclusion when $L^\prime \neq 1$ where $L^\prime = \lim\limits_{n \to \infty} b_n$ with $b_n = n \left(\frac{a_n}{a_{n+1}} - 1\right)$. The existence of $L^\prime$ implies that $L = \lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$ exists. This is exactly the case where ratio test is not conclusive.


However, in that case Raabe test can be conclusive. For example Raabe test proves that $\sum \frac{1}{n^2}$ is convergent while ratio test cannot provide a conclusion. See link to Wikipedia for the detailed definition of Raabe test.
