Sum series- convergence and divergence

[Chipset3600]

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]

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]

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 don't understood...
Why u said that : \(\displaystyle \lambda_{n} = \ln a_{n}\) ?

 

[chisigma]

I don't 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]

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

Kind regards

$\chi$ $\sigma$

Still don't understanding, and i never study "difference equation".

 

[chisigma]

Still don't 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]

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

 

[chisigma]

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

If You intend to improve Your knowledge about difference equations, on MHB there is a tutorial thread...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html

Kind regards

$\chi$ $\sigma$

 

[Evgeny.Makarov]

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]

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]

Other way to solve it:

Spoiler

http://i.imgur.com/8gkjO4Z.jpg

 

[zzephod]

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).

.