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


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 $\sum a_{n}$.


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?

  • $\begingroup$ Let $a_{2n}=\frac{1}{3^n}$ and $a_{2n+1}=\frac{2}{3^n}.$ Then your limsup is $3.$ $\endgroup$ May 15 at 15:26
  • $\begingroup$ 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. $\endgroup$ May 15 at 15:33
  • $\begingroup$ @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 ($\frac{2}{3^{n}}$?) $\endgroup$
    – Debu
    May 15 at 15:35
  • $\begingroup$ Yes, sorry, came up with a simpler example but failed to change the limsup value at the end. $\endgroup$ May 15 at 15:38
  • $\begingroup$ 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. $\endgroup$ May 15 at 15:39

1 Answer 1


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 $\limsup\left\lvert \frac{a_{n+1}}{a_{n}} \right\rvert =\infty$.


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