# Give a sequence such that root test works while ratio test fails

Question: Give a sequence $(a_n)_{n=1}^\infty$ with $a_n>0$ such that root test works while ratio test does not work, that is, $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} \text{ exists}$$ while $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} \text{ does not exist}.$$

My attempt:

Define a sequence $(a_n)_{n=1}^\infty$ such that $$a_n = \begin{cases} 2^{n-1} & \text{if }n \text{ is odd,} \\ 2^{n+1} & \text{if }n \text{ is even}. \end{cases}$$
Odd subsequence $(a_{n_j})_{j=1}^\infty$ implies that $\lim_{j\rightarrow\infty}(a_{n_j})^{\frac{1}{n_j}} = \lim_{j\rightarrow\infty} 2^{1-\frac{1}{n_j}} = 2$ while even subsequence $(a_{n_k})_{k=1}^\infty$ implies that $\lim_{k\rightarrow\infty}(a_{n_k})^{\frac{1}{n_k}} = \lim_{k\rightarrow\infty} 2^{1+\frac{1}{n_k}} = 2.$ It follows that $\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} = 2,$ that is, root test works. However, for odd subsequence $(a_{n_j})_{j=1}^\infty,$ $$\lim_{j\rightarrow\infty}\frac{a_{n_j+1}}{a_{n_j}} = \lim_{n\rightarrow\infty} \frac{2^{n_j+1}}{2^{n_j-1}} = \lim_{n\rightarrow\infty} 4 = 4.$$ For even subsequence $(a_{n_k})_{k=1}^\infty,$ $$\lim_{k\rightarrow\infty}\frac{a_{n_k+1}}{a_{n_k}} = \lim_{k\rightarrow\infty} \frac{2^{n_k-1}}{2^{n_k+1}} = \lim_{k\rightarrow\infty} \frac{1}{4} = \frac{1}{4}.$$
Therefore, $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ does not exist, that is, ratio test does not work.

Does my example work?

• Yes it is right Commented Aug 16, 2017 at 3:00

Your example is right. You can construct a similar exemple such that the series $\sum\limits_{n=1}^{\infty} a_n$ converges as follows: $$a_n = \begin{cases} \frac{1}{2^{n+1}} & \text{if n is odd,} \\ \frac{1}{2^{n}} & \text{if n is even}. \end{cases}$$ Then for the odd subsequence $(a_{n_j})_{j=1}^\infty$ we have $\lim\limits_{j\rightarrow\infty}(a_{n_j})^{\frac{1}{n_j}} = \lim\limits_{j\rightarrow\infty} \frac{1}{2^{1+\frac{1}{n_j}}}= \frac{1}{2}$ and for the even subsequence $(a_{n_k})_{k=1}^\infty$ we have $\lim\limits_{k\rightarrow\infty}(a_{n_k})^{\frac{1}{n_k}} = \lim\limits_{k\rightarrow\infty} \frac{1}{2}=\frac{1}{2}$, this implies that $(a_n)^{\frac{1}{n}} \to \frac{1}{2}$, therefore the root test works and the series converges absolutely.
But for odd $n$, $$a_n = \frac{1}{2^{n+1}}= a_{n+1}.$$ So, if the limit $\lim\limits_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$ exists, it must be equal to $1$, which makes the ratio test inconclusive. Actually, the limit does not exist, because for even $n$, $\frac{a_{n+1}}{a_n} = \frac{1}{4}$.