# $a_1=1/3$, $a_{n+1}=\sqrt[n]{a_n}$, then does $\sum{a_n}$ converge?

Question_

The sequence $$\{a_n\}$$ is given by $$a_1=1/3$$ and $$a_{n+1}=\sqrt[n]{a_n}$$. then does $$\sum_{n=1}^\infty{a_n}$$ converge?

I've tried to use several tests, such as the ratio test or the root test. When we use the ratio test: $$\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty}\left|\frac{a_n^{1\over n}}{a_n}\right|=\lim_{n \to \infty}\left|{a_n^{\frac{1-n}{n}}}\right|$$ This is not a good method I think. It was not that hard to derive the closed-form of the sequence: $$a_n=\left({1\over3}\right)^{1\cdot{1\over2}\times\cdots\times{1\over{n-1} }}=\left({1\over3}\right)^{1\over (n-1)!}$$ Using this, I re-tried the ratio test: $$\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty}\left|\left({1\over3}\right)^{{1\over n!}-{1\over (n-1)!}}\right|=\lim_{n \to \infty}\left|\left({1\over3}\right)^{1-n\over n!}\right|=\left|\left({1\over3}\right)^0\right|=1$$ If we get $$1$$, we cannot conclude whether $$\sum a_n$$ converges or diverges. So, in this time, I applied the root test: $$\lim_{n \to \infty}\sqrt[n]{a_n}=\lim_{n \to \infty}\left|\left({1\over3}\right)^{{1\over n!}}\right|=\left|\left({1\over3}\right)^0\right|=1$$ Again I got $$1$$, which is the value that we cannot also conclude whether it converges or diverges. There exist other method but I could not try beacuse

1. in the case of the integral test, I cannot ensure that $$f(x)=\left({1\over3}\right)^{1\over (x-1)!}$$ is a decreasing-function, and even so, I cannot integrate $$f(x)$$.

2. Setting the other sequences $$\{b_n\}$$ and finding $$\lim_{n \to \infty}\frac {a_n}{b_n}$$ can be also a suitable method. However, it was difficult to put a fitable $$\{b_n\}$$.

Could you please suggest other great methods to solve the problem? Thanks.

• Your $a_n \to 1$ as $n \to \infty$. But the terms of a convergent series must go to $0$. Feb 19, 2020 at 3:41
• Thanks!! I was stupid,,,
– ToBY
Feb 19, 2020 at 3:42

Note that $$\sqrt[n]{a_n}$$ in fact grows with $$n$$, because $$a_i \leq 1$$ for all $$i$$ can be easily seen by induction, and then $$a_{n+1} = \sqrt[n]{a_n} \geq a_n$$ (because the $$n$$th root function has this property between $$0$$ and $$1$$). So each $$a_i$$ is bigger than $$a_1 = \frac 13$$ : for merely this reason alone, the sequence does not converge.
This is also clear from your closed form : as $$n$$ increases, $$\frac 1{(n-1)!}$$ decreases, so tends to zero : by continuity of $$x \to a^x$$ ($$a = 1/3$$) the series term converges to $$1$$, so there's no convergence. The ratio test will fail because the ratio is also converging to $$1$$.