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Sum Series:

Chipset3600

Member
Feb 14, 2012
79
Hello MHB,
How can i study the convergence or divergence of this serie:
From: Problems and Excercises of Analysis Mathematic- B. Demidovitch (nº: 2446):

[TEX]\frac{2}{1}+\frac{2.5.8}{1.5.9}+\frac{2.5.8.11.14..(6n-7).(6n-4)}{1.5.9.13.17...(8n-11).(8n-7)}+...[/TEX]
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Have you tried the ratio test ?
 

Chipset3600

Member
Feb 14, 2012
79

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We need to calculate

\(\displaystyle \lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|\)

So what is $a_n$ ?
 

Chipset3600

Member
Feb 14, 2012
79
We need to calculate

\(\displaystyle \lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|\)

So what is $a_n$ ?
Don't know how to find this [TEX]a_n[/TEX]
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$a_n$ is the general term ,specifically, the term that contains n , can you find it in your series ?

For example

\(\displaystyle S=1+2+3+\cdots +n+\cdots \)

Then $a_n= n$

and

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
By the way , I think your series should be

\(\displaystyle \frac{2}{1}+\frac{2\cdot 5}{1\cdot 5 }+\frac{2\cdot 5 \cdot 8}{1\cdot 5 \cdot 9}+\cdots+\frac{2\cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdots (6n-7).(6n-4)}{1\cdot 5 \cdot 9 \cdot 13 \cdot 17 \cdots (8n-11) \cdot(8n-7)}+\cdots\)
 

Chipset3600

Member
Feb 14, 2012
79
By the way , I think your series should be

\(\displaystyle \frac{2}{1}+\frac{2\cdot 5}{1\cdot 5 }+\frac{2\cdot 5 \cdot 8}{1\cdot 5 \cdot 9}+\cdots+\frac{2\cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdots (6n-7).(6n-4)}{1\cdot 5 \cdot 9 \cdot 13 \cdot 17 \cdots (8n-11) \cdot(8n-7)}+\cdots\)
maybe, but not so in the book :/
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

chisigma

Well-known member
Feb 13, 2012
1,704
Hello MHB,
How can i study the convergence or divergence of this serie:
From: Problems and Excercises of Analysis Mathematic- B. Demidovitch (nº: 2446):

[TEX]\frac{2}{1}+\frac{2.5.8}{1.5.9}+\frac{2.5.8.11.14..(6n-7).(6n-4)}{1.5.9.13.17...(8n-11).(8n-7)}+...[/TEX]
It is easy to see that the general term of the series obeys to the difference equation...


$\displaystyle a_{n+1} = a_{n}\ \frac{(6 n - 7)\ (6 n - 4)}{(8 n - 11)\ (8 n - 7)}\ (1)$

... so that is...

$\displaystyle \frac {a_{n+1}}{a_{n}} = \frac{(6 n - 7)\ (6 n - 4)}{(8 n - 11)\ (8 n - 7)} = \frac{(6 - \frac{7}{n})\ (6 - \frac{4}{n})}{(8 - \frac{11}{n})\ (8 - \frac{7}{n})}\ (2)$

... and because is...

$\displaystyle \lim_{n \rightarrow \infty} a_{n}=0\ (3)$

... and...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}= \frac{9}{16} < 1\ (4)$

... the series converges...

Kind regards

$\chi$ $\sigma$
 

Chipset3600

Member
Feb 14, 2012
79
It is easy to see that the general term of the series obeys to the difference equation...


$\displaystyle a_{n+1} = a_{n}\ \frac{(6 n - 7)\ (6 n - 4)}{(8 n - 11)\ (8 n - 7)}\ (1)$

... so that is...

$\displaystyle \frac {a_{n+1}}{a_{n}} = \frac{(6 n - 7)\ (6 n - 4)}{(8 n - 11)\ (8 n - 7)} = \frac{(6 - \frac{7}{n})\ (6 - \frac{4}{n})}{(8 - \frac{11}{n})\ (8 - \frac{7}{n})}\ (2)$

... and because is...

$\displaystyle \lim_{n \rightarrow \infty} a_{n}=0\ (3)$

... and...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}= \frac{9}{16} < 1\ (4)$

... the series converges...

Kind regards

$\chi$ $\sigma$
thank you, I would not get the response so soon!