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Sum series- convergence and divergence

Chipset3600

Member
Feb 14, 2012
79
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
Feb 13, 2012
1,704
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
Feb 14, 2012
79
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
Feb 13, 2012
1,704
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
Feb 14, 2012
79
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
Feb 13, 2012
1,704
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
Feb 14, 2012
79
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

Well-known member
Feb 13, 2012
1,704

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
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
Feb 14, 2012
79
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
Feb 14, 2012
79

zzephod

Well-known member
Feb 3, 2013
134
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).

.