TITLE: Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$
QUESTION [8 upvotes]: Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$

My thoughts:
By the theorem: Suppose $a_n\ge0$ for all $n$, and let $l=\displaystyle\limsup{\sqrt[n]{a_n}}$. If $l<1$, then $\displaystyle\sum_{n=1}^\infty a_n$ converges; and if $l>1$, then $\displaystyle\sum_{n=1}^\infty a_n$ diverges.
Thus, we know $\displaystyle\sum_{n=1}^\infty {a_{n+1}\over{a_n}}$ diverges
I guess it might works, $(x_n)_{n=1}^{+\infty}$, $x_{n+1} = x_n^2 + x_n$ for all $n\ge1$..

REPLY [7 votes]: I have an example:
Consider
$$ a_n = \begin{cases} \frac{1}{m^2} & n = 2m - 1\\ \frac{1}{2^m} & n = 2m \end{cases}\qquad m = 1, 2, \ldots.  $$
It's obvious that the series $\sum a_n$ is convergent since $\sum \frac{1}{n^2}$ and  $\sum \frac{1}{2^n}$ are both convergent.
However,
$$\limsup \frac{a_{n+1}}{a_n}=\lim \frac{2^n}{n^2}=\infty.$$