3
$\begingroup$

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.

$\endgroup$
2
  • 4
    $\begingroup$ Your $a_n \to 1$ as $n \to \infty$. But the terms of a convergent series must go to $0$. $\endgroup$ Commented Feb 19, 2020 at 3:41
  • $\begingroup$ Thanks!! I was stupid,,, $\endgroup$
    – ToBY
    Commented Feb 19, 2020 at 3:42

1 Answer 1

5
$\begingroup$

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 $1$.

Moral : Check the convergence of individual terms to zero, before applying any test.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .