# Sum Series:

#### Chipset3600

##### Member
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
Have you tried the ratio test ?

#### Chipset3600

##### Member
Have you tried the ratio test ?
Is not a geometric serie. How can i use the ratio test in limit without know the [TEX]a_{}n[/TEX] ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
We need to calculate

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

So what is $a_n$ ?

#### Chipset3600

##### Member
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
$a_n$ 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
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
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 :/

MHB Math Helper

#### chisigma

##### Well-known member
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
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!