# Sum series- convergence and divergence

#### Chipset3600

##### Member
converge or diverge?

$$\displaystyle \sum_{n=1}^{^{\infty }}a_{n}$$

$$\displaystyle a_{1}= \frac{1}{3}, a_{n+1}= \sqrt[n]{a_{n}}$$
Im having problems to solve this exercise, i would like to see your solutions

#### chisigma

##### Well-known member
converge or diverge?

$$\displaystyle \sum_{n=1}^{^{\infty }}a_{n}$$

$$\displaystyle a_{1}= \frac{1}{3}, a_{n+1}= \sqrt[n]{a_{n}}$$
Im having problems to solve this exercise, i would like to see your solutions
Defining $\displaystyle \lambda_{n} = \ln a_{n}$ the difference equation becomes...

$\displaystyle \lambda_{n+1} = \frac{\lambda_{n}}{n}\ (1)$

... and for any $\displaystyle a_{1} \ne 0$ is $\displaystyle \lim_{n \rightarrow \infty} \lambda_{n}=0$, so that $\displaystyle \lim_{n \rightarrow \infty} a_{n}=1$, so that...

Kind regards

$\chi$ $\sigma$

#### Chipset3600

##### Member
Defining $\displaystyle \lambda_{n} = \ln a_{n}$ the difference equation becomes...

$\displaystyle \lambda_{n+1} = \frac{\lambda_{n}}{n}\ (1)$

... and for any $\displaystyle a_{1} \ne 0$ is $\displaystyle \lim_{n \rightarrow \infty} \lambda_{n}=0$, so that $\displaystyle \lim_{n \rightarrow \infty} a_{n}=1$, so that...

Kind regards

$\chi$ $\sigma$
I dont understood...
Why u said that : $$\displaystyle \lambda_{n} = \ln a_{n}$$ ?

#### chisigma

##### Well-known member
I dont understood...
Why u said that : $$\displaystyle \lambda_{n} = \ln a_{n}$$ ?
It is the first step to demonstrate that $\lim_{n \rightarrow \infty} a_{n}=1$...

Kind regards

$\chi$ $\sigma$

#### Chipset3600

##### Member
It is the first step to demonstrate that $\lim_{n \rightarrow \infty} a_{n}=1$...

Kind regards

$\chi$ $\sigma$
Still dont understanding, and i never study "difference equation".

#### chisigma

##### Well-known member
Still dont understanding, and i never study "difference equation".
Yourself wrote...

$\displaystyle a_{1}= \frac{1}{3},\ a_{n+1} = \sqrt[n] {a_{n}}\ (1)$

That is a 'difference equation' the solution of which is $\displaystyle a_{n}$...

Kind regards

$\chi$ $\sigma$

#### Chipset3600

##### Member
Yourself wrote...

$\displaystyle a_{1}= \frac{1}{3},\ a_{n+1} = \sqrt[n] {a_{n}}\ (1)$

That is a 'difference equation' the solution of which is $\displaystyle a_{n}$...

Kind regards

$\chi$ $\sigma$
Im really trying to understand, but sorry!
resolution is too advanced for me to realize

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
chisigma's suggestion is to see what happens to the original equation $a_{n+1}= \sqrt[n]{a_{n}}$ when we make a definition $\lambda_n=\ln(a_n)$. Then $\lambda_{n+1}\stackrel{\text{def}}{=} \ln(a_{n+1})= \ln((a_n)^{1/n}) =\ln(a_n)/n \stackrel{\text{def}}{=}\lambda_n/n$ Thus, $\lambda_n\to0$ as $n\to\infty$, and $a_n=e^{\lambda_n}\to e^0=1$ as $n\to\infty$.

#### Chipset3600

##### Member
The definition used is that it was strange for me...
My teacher give me other solution.
Taking the limit it will be equal to 1.
$$\displaystyle a_{n} = (\frac{1}{3})^{\frac{1}{n!}}$$

#### Chipset3600

##### Member
Other way to solve it:

#### zzephod

##### Well-known member
converge or diverge?

$$\displaystyle \sum_{n=1}^{^{\infty }}a_{n}$$

$$\displaystyle a_{1}= \frac{1}{3}, a_{n+1}= \sqrt[n]{a_{n}}$$
Im having problems to solve this exercise, i would like to see your solutions
A quick numerical experiment shows that $$\displaystyle a_n \to 1$$ and if this is indeed the case the series $$\displaystyle \sum_{n=1}^{^{\infty }}a_{n}$$ diverges since to converge $$\displaystyle a_n$$ must go to zero.

The simplest method of proving that $$\displaystyle a_n \not\to 0$$ is to show that if $$\displaystyle 0<a_1<a_k<1$$ then $$\displaystyle 0<a_1<a_{k+1}<1$$ which will show that $$\displaystyle a_n$$ is bounded below by $$\displaystyle a_1$$.

The approach I would use is induction based on the observation that if $$\displaystyle 0<a_k<1$$ and $$\displaystyle a_{k+1}=\sqrt[k]{a_{k}}$$ then:

$$\displaystyle a_{k+1} < 1$$

and

$$\displaystyle a_{k+1}^k=a_k$$

but for any $$\displaystyle \kappa>1$$ we now can conclude $$\displaystyle a_{k+1}^{\kappa}<a_{k+1}$$ and so:

$$\displaystyle a_k<a_{k+1}$$

Thus $$\displaystyle \{a_n\}$$ is an increasing sequence and so bounded below by its first term.

In addition any other method that shows that $$\displaystyle \{a_n\}$$ is an non-decreasing sequence would do (which includes at least one form of the ratio test).

.