# Convergence when ratio test=1

When using the ratio test for absolute convergence of a series $$\sum_{n=1}^\infty a_{n}$$, if the limit of the ratio $$|a_{n+1}|/|a_{n}|=1$$ when $$n \rightarrow \infty$$, the fate of the series is indeterminate. However, if $$|a_{n+1}|\ge|a_{n}|$$ for all sufficiently large values of $$n$$, does that imply that the series is divergent?

For instance, if I am correct, the series $$\sum_{n=1}^\infty (n!/n^n)x^n$$ converges for $$|x| and diverges if $$|x|>e$$.

But, when $$|x|=e$$, the limit of the ratio $$=1$$ and $$|a_{n+1}|=|a_{n}|e(n/(n+1))^n$$ where $$(n/(n+1))^n>1/e$$ for all values of $$n$$, so that $$|a_{n+1}|\ge|a_{n}|$$, and the series is then divergent. Is this correct?

On the other hand, if $$|a_{n+1}|<|a_{n}|$$ for large values of $$n$$, we cannot conclude, is this true?

• If $|a_{n + 1}| \ge |a_n|$ for all large values of $n$, then either a) the terms are eventually all zero or b) the terms don't tend to zero in magnitude, and the series is divergent. – user296602 Feb 26 at 20:24

If $$\vert a_{n+1} \vert \geq \vert a_n\vert$$ when $$n>N$$, then for all sufficiently large $$n$$, $$\vert a_n \vert$$ is greater than a constant $$a_N$$. Thus, $$\lim_{x\to\infty}a_n\not=0$$. (Unless in the trivial case that $$a_n=0$$ for all sufficiently large $$n$$.) Hence $$a_n$$ diverges.
If $$\vert a_{n+1} \vert \lt \vert a_n\vert$$ for large $$n$$, then we cannot conclude. For instance, $$\sum\frac{1}{n}$$ diverges, but $$\sum\frac{1}{n^2}$$ converges.