0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user296602 Feb 26 at 20:24
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.