What are the conditions of the test of reason does not apply to a series ?
I have no idea at all what you mean by "test of reason"!
Is this a language problem? Do you mean "ratio test"?
(Hey, makes sense- ratio- rational- reason!)
The ratio test for convergence of a series of positive numbers, $\displaystyle \sum_{n=0}^\infty a_n$, is to look at the limit of the ration $\displaystyle \frac{a_{n+1}}{a_n}$. If that limit is less than 1, the series converges. If it is greater than 1, the series converges. If it equal to one, it does not tell us if the series converges or not.
So one situation in which the ratio test does not work (I am not sure I would say "does not apply") is when that limit of the ratio is 1.
On the face of it, it looks like the ratio test can only apply to series of positive numbers, but if you have a series of all negative numbers, you could factor out -1 and apply it to the new series with positive numbers which converges if and only if the original series does.
If you have a mixture of positive and negative numbers, you cannot apply the ratio test- although you could take the absolute value and determine absolute convergence. If a series converges absolutely, then it converges but there exist many series which converge, but not absolutely.
Thank you. Sorry about the confusion with the language
Problem:
Verify that the ratio test can be applied to investigate the convergence of each series. If so apply it, if not explain why the test can not be applied
1) $\displaystyle \sum_{n=1}^{\infty} \frac{n}{3^n}$
My solution:
$\displaystyle \frac{a_{n+1}}{a_n} = \frac{(n+1)}{3^{n+1}} . \frac{3^n}{n} = \frac{n+1}{3n}$
$\displaystyle \lim_{n\to{\infty}} \frac{n+1}{3n} = \frac{1}{3}$
Convergent
It is correct ?
2) $\displaystyle \sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}$
How do I calculate this?
The first is correct. For the second
$\displaystyle
\lim_{n \to \infty} \frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^ 3+1}} = 1.
$
To drawn conclusions from the ratio test, if
$\displaystyle
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L
$
then $\displaystyle L < 1 $ converent and if $\displaystyle L > 0$ divergent. No conclusion for $\displaystyle L = 1$.
$\displaystyle
\frac{\frac{(n+1)^2+1}{(n+1)^3+1}}{\frac{n^2+1}{n^ 3+1}} =
\frac{(n+1)^2+1}{n^2+1} \cdot \frac{n^3+1}{(n+1)^3+1} = \frac{n^2+2n + 2}{n^2+1} \cdot \frac{n^3+1}{n^3 + 3n^2 + 3n + 2}
$
$\displaystyle
= \frac{1+\frac{2}{n} + \frac{2}{n^2}}{1 + \frac{1}{n^2}} \cdot \frac{1 + \frac{1}{n^3}}{1 + \frac{3}{x} + \frac{3}{n^2} + \frac{2}{n^3}}
$
as $\displaystyle n \to \infty $ everthing drops except for the 1's.