# Is there a sequence $a_n$ for which $\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n}|$ doesn't exist, but $\sum a_n$ is convergent?

The question is in the title:

Is there a sequence $a_n$ for which:

1. $\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n}|$ doesn't exist (so it's not defined, which means we're looking for a case when it's now even $+\infty$ nor $-\infty)$

2. $\sum_{n=1}^{+\infty} a_n$ is convergent

According to my workbook, such a sequence exists but I cannot find it. Any hints?

$$\forall n\geqslant1,\qquad a_{2n}=a_{2n-1}=2^{-n}$$ Edit: To see that the ratios $a_{n+1}/a_n$ can have $0$ and $+\infty$ as limit points simultaneously, consider $$\forall n\geqslant1,\qquad a_{2n}=2^{-n},\quad a_{2n-1}=3^{-n}$$

Take for example the sequence

$$a_n=\frac{\sin n}{n^2}$$

Take $a_n=0$. The ratio is indeterminate and does not exist.

Another example: $$a_n=\frac{(2+(-1)^n)^n}{3^nn^2}.$$ Partial limits of $\frac{a_{n+1}}{a_n}$ are $\frac13$ and $1$, but $a_n\leq\frac{1}{n^2}$, so $\sum a_n$ converges.

• Partial limits (of the ratio) are 0 and +oo. – Did Dec 7 '14 at 8:17

There are different types of non-existent limits (divergent limits). Some are infinitely divergent, and some simply fail to converge. Still others converge to different values depending on direction of approach.

I would have to say that infinite divergent ratios make a sum divergent. As does any convergent ratio larger than 1. Any ratio limit, convergent or divergent, that is larger than 1, proves the sum is divergent.

As far as infinite series go, there is only one direction from which to approach the sum.

BUT... if the ratio simply fails to converge - such as due to oscillations - then I do not believe this says anything about the series.