# $\limsup_{n \to \infty} \Big|\frac{a_{n+1}}{a_{n}}\Big| > 1$ does not imply the divergence of $\sum a_{n}$

Question

$$\limsup_{n \to \infty} \Big|\frac{a_{n+1}}{a_{n}}\Big| > 1$$ does not imply the divergence of $$\sum a_{n}$$.

Context

In baby rudin, chapter three, theorem3.34 says:-

a.) If $$\limsup_{n \to \infty} \Big|\frac{a_{n+1}}{a_{n}}\Big| < 1 \implies \sum a_{n}$$ converges.

b.) If $$\Big|\frac{a_{n+1}}{a_{n}}\Big| \ge 1$$ for $$n \ge n_0$$ $$\implies \sum a_{n}$$ diverges.

There is an asymmetry in the above theorem, which wasn't there in root test. So I am looking for a counter example where $$\limsup_{n \to \infty} \Big|\frac{a_{n+1}}{a_{n}}\Big| > 1$$ does not imply the divergence of $$\sum a_{n}$$.

Attempt

Does the following example work:-

Let $$a_{n} = \{\frac{1}{2^{3}},\frac{1}{2^{2}},\frac{1}{4^{3}},\frac{1}{4^{2}},\frac{1}{6^{3}},\frac{1}{6^{2}} \ldots\}$$. We note that $$a_{n} \leq \frac{1}{n^{2}}$$. Hence by comparison test $$a_{n}$$ converges and infact $$\limsup_{n \to \infty}\Big|\frac{a_{n+1}}{a_{n}}\Big| = \infty$$

Could somebody verify if this is fine?

• Let $a_{2n}=\frac{1}{3^n}$ and $a_{2n+1}=\frac{2}{3^n}.$ Then your limsup is $3.$ May 15 at 15:26
• Basically, the $\limsup>1$ requires infinitely many $n$ such that $|a_{n+1}|>|a_n|.$ But we can make those $n$ sparse enough to make the series converge. (Technically, that is only a necessary condition for $\limsup |a_{n+1}/a_n|\geq 1,$ to make it $>1,$ we need a little more. May 15 at 15:33
• @ThomasAndrews In the example you provided the limsup is 2, right? Could you help me see quickly how the series converges? It seems like I need to compare it with some geometric series ($\frac{2}{3^{n}}$?)
– Debu
May 15 at 15:35
• Yes, sorry, came up with a simpler example but failed to change the limsup value at the end. May 15 at 15:38
• We can even find examples where $\limsup |a_{n+1}/a_n|=+\infty.$ Let $a_{2n}=\frac1{n^3}, a_{2n+1}=\frac{1}{n^2},$ for example. May 15 at 15:39

Take any absolutely convergent series $$\sum \alpha_m$$ and any sequence $$(r_m)$$ with $$r_n \ge 1$$. Then also $$\sum \alpha_m/r_m$$ is absolutely convergent. Define $$a_n = \begin{cases} \alpha_m/r_m & n = 2m \\ \alpha_m & n = 2m+1 \end{cases}$$ Then $$\sum a_n$$ converges absolutely and $$\left\lvert \frac{a_{2m+1}}{a_{2m}} \right\rvert = r_m .$$ Thus $$\limsup\left\lvert \frac{a_{n+1}}{a_{n}} \right\rvert \ge \limsup r_m .$$ Now take your favorite sequence $$(r_m)$$ with $$\limsup r_m > 1$$. An eaxmple is $$r_m = m$$ in which case $$\limsup\left\lvert \frac{a_{n+1}}{a_{n}} \right\rvert =\infty$$.