Convergence/Divergence of Series 2446

[Chipset3600]

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

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

 

[alyafey22]

Have you tried the ratio test ?

 

[Chipset3600]

Have you tried the ratio test ?

Is not a geometric serie. How can i use the ratio test in limit without know the $$a_{}n$$ ?

 

[alyafey22]

We need to calculate

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

So what is $a_n$ ?

 

[Chipset3600]

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 $$a_n$$

 

[alyafey22]

$a_n$ 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

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

 

[alyafey22]

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]

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 :/

 

[alyafey22]

maybe, but not so in the book :/

Strange if so .

 

[chisigma]

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

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

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]

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!