# Example of a series where the ratio test $\limsup |a_{n+1}/a_n|$ can be applied, but $\lim |a_{n+1}/a_n|$ cannot

The ratio test asserts the absolute convergence of $$\sum_{n\geq 1}a_n$$ if $$\limsup \bigg |\frac{a_{n+1}}{a_n}\bigg |<1$$ In calculus, we learn the seemingly weaker form $$\lim \bigg |\frac{a_{n+1}}{a_n}\bigg |<1$$ Similarly for the root test. Is there an example that shows the $$\limsup$$ version is actually stronger? I am looking for an example where the weaker version cannot be used at all, or it is not obvious how to use the weaker version.

To give an idea of what I am NOT looking for, we learn that the root test is strictly stronger than the ratio test, in the sense that the root test implies the ratio test, and that there are series for which the ratio test fails, but the root test succeeds. An example is given by $$\frac{1}{2}+1+\frac{1}{8}+\frac{1}{4}+\frac{1}{32}+\frac{1}{16}+\frac{1}{128}+\frac{1}{64}+\dots$$ However, once we rewrite the above as $$\bigg(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots\bigg)+\bigg(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\dots\bigg)$$ After barely any work, the ratio test now applies, making for an unsatisfying example of the root test being strictly stronger than the ratio test.

• It is nor clear what you mean by the root test implies the ration test. It is known that $\liminf_n\frac{|a_{n+1}|}{|a_n|}\leq\liminf_n\sqrt[n]{|a_n|}\leq\limsup_n\sqrt[n]{|a_n|}\leq\limsup_n\frac{|a_{n+1}|}{|a_n|}$. Hence if the ratio test works, so does the root test. There are cases, however, where the root test works but the ratio test fails. Commented Jul 19 at 15:33

Consider the series $$a_n$$ obtained as follows

• $$a_1=1$$,
• $$a_2=1$$, $$a_3=2$$
• $$a_4=1$$, $$a_5=2$$, $$a_6=3$$
• $$a_7=1$$, $$a_8=2$$, $$a_9=3$$, $$a_{10}=4$$, etc.

That is, split $$\mathbb{N}$$ in blocks $$A_k$$, each of size $$k$$. On $$A_k$$, the sequence $$a$$ is monotone increasing and takes values in $$\{1,\ldots,k\}$$. Thus, for $$n\geq1$$ $$a_{1+\tfrac{n(n+1)}{2}}=1,\,a_{1+\tfrac{n(n+1)}{2}+1}=2, \ldots ,a_{1+\frac{n(n+2)}{2}+n}=n$$

Consider the series $$\sum_n b_n$$ where $$b_n=4^{-n}a_n$$. Observe that

$$\liminf_m\frac{a_{m+1}}{a_m}=\lim_n\frac14\frac{1}{n}=0$$ whilst $$\limsup_m\frac{a_{m+1}}{a_m}=\frac12<1$$

It is not difficult to show that $$\lim_m\sqrt[m]{a_m}=\frac14$$.