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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 $|x|<e$ 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?

user460753
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  • 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. –  Feb 26 '19 at 20:24

1 Answers1

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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.

fEst1ck
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