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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

\(\displaystyle \sum_{n=1}^{^{\infty }}a_{n} \)

\(\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

- Thread starter Chipset3600
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- Thread starter
- #1

- Feb 14, 2012

- 79

\(\displaystyle \sum_{n=1}^{^{\infty }}a_{n} \)

\(\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

- Feb 13, 2012

- 1,704

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

\(\displaystyle \sum_{n=1}^{^{\infty }}a_{n} \)

\(\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

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

... 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

$\chi$ $\sigma$

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- #3

- Feb 14, 2012

- 79

I dont understood...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

$\chi$ $\sigma$

Why u said that : \(\displaystyle \lambda_{n} = \ln a_{n}\) ?

- Feb 13, 2012

- 1,704

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

Why u said that : \(\displaystyle \lambda_{n} = \ln a_{n}\) ?

Kind regards

$\chi$ $\sigma$

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- #5

- Feb 14, 2012

- 79

Still dont understanding, and i never study "difference equation".It is the first step to demonstrate that $\lim_{n \rightarrow \infty} a_{n}=1$...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

Yourself wrote...Still dont understanding, and i never study "difference equation".

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

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

Kind regards

$\chi$ $\sigma$

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- #7

- Feb 14, 2012

- 79

Im really trying to understand, but sorry!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

$\chi$ $\sigma$

resolution is too advanced for me to realize

- Feb 13, 2012

- 1,704

If You intend to improve Your knowledge about difference equations, on MHB there is a tutorial thread...Im really trying to understand, but sorry!

resolution is too advanced for me to realize

http://mathhelpboards.com/discrete-...rence-equation-tutorial-draft-part-i-426.html

Kind regards

$\chi$ $\sigma$

- Jan 30, 2012

- 2,528

- Thread starter
- #10

- Feb 14, 2012

- 79

My teacher give me other solution.

Taking the limit it will be equal to 1.

\(\displaystyle a_{n} = (\frac{1}{3})^{\frac{1}{n!}}\)

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- #11

- Feb 14, 2012

- 79

Other way to solve it:

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.

\(\displaystyle \sum_{n=1}^{^{\infty }}a_{n} \)

\(\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

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).

.