1. ## Convergence of Series

Good evening All,

I am wrapping up my final problem in the Root & Ratio Test section. I have the following:

$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{2n-1}{3n-2}$

So...

$\displaystyle \displaystyle\lim_{n\rightarrow\infty}|\frac{2(n+1 )-1}{3(n+1)-1}*\frac{3n-2}{2n-1}|$

So...

$\displaystyle \displaystyle\lim_{n\rightarrow\infty}|\frac{2n+1} {3n+2}*\frac{3n-2}{2n-1}|$

So...

$\displaystyle \displaystyle\lim_{n\rightarrow\infty}|\frac{6n^2-n-2}{6n^2+n-2}|=1$

Inconclusive by the Ratio Test...

What's the next plan?

If I apply the Divergence Test (which should probably have been the first thing I did), I find...

$\displaystyle \displaystyle\lim_{n\rightarrow\infty}\frac{2n-1}{3n-2} = \frac{2}{3} \neq 0$

Therefore, by the Divergence Test the series Diverges

Does anyone agree or disagree with this?

2. The first line of yours is not true at all!

3. $\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{2n-1}{3n-2}$

Could you please explain? I thought that the two meant essentially the same thing.

4. Compute to n=1,2,3,4 instead to inf, in both sides...

5. If we write the series as...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)} = \sum_{n=1}^{\infty} a_{n}$ (1)

... the DE defining the $\displaystyle a_{n}$ is...

$\displaystyle \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1$ (2)

From (2) it follows that is...

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

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} = \frac{2}{3} < 1$ (3)

... so that the series (1) converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Cauchy convergence criteria:

$\displaystyle \displaystyle \sum_{n=1}^{\infty}a_n$

$\displaystyle \displaystyle q= \lim_{n\to\infty} \sqrt[n]{a_n}$

$\displaystyle \displaystyle q=\begin{cases} q<1& converges \\ q>1& diverges \\ q=1 & ? need \;more\; testing \end{cases}$

D'alambert convergence criteria :

$\displaystyle \displaystyle \sum_{n=1}^{\infty}a_n$

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

$\displaystyle \displaystyle q=\begin{cases} q<1& converges \\ q>1& diverges \\ q=1 & ? need\; more\; testing \end{cases}$

Raabel's convergence criteria:

$\displaystyle \displaystyle \sum_{n=1}^{\infty}a_n$

$\displaystyle \displaystyle q= \lim_{n\to\infty} n[ \frac{a_{n+1}}{a_n} -1]$

$\displaystyle \displaystyle q=\begin{cases} q>1& converges \\ q<1& diverges \\ q=1 & ? need\; more\; testing \end{cases}$

for alternative series u'll use
Leibnitz convergence criteria:

$\displaystyle \displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}a_n$

converges if the conditions are met that :

1° $\displaystyle \displaystyle \lim_{n\to\infty} a_n = 0$

2° $\displaystyle a_n$ monotonically decreasing

if not then it diverges

or u can test it by the definition that say that $\displaystyle \displaystyle \sum_{n=1}^{\infty}a_n$ converges if $\displaystyle \displaystyle S_N= \sum_{n=1}^{\infty}a_n=a_1+a_2+ . . . + a_N$ sequence $\displaystyle (S_N)_{N\in\mathbb{N}}$ which is sequence of partial sum, is :

$\displaystyle \displaystyle \lim_{N\to\infty}S_N =\begin{cases} final \;number\Rightarrow \;series \;converges \\ infinite\; or\; does \;not \;exist\Rightarrow \;series\; diverges \end{cases}$

there are more criteria but, it's not for real numerical series... it's for function series ...

7. Originally Posted by chisigma
If we write the series as...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)} = \sum_{n=1}^{\infty} a_{n}$ (1)

... the DE defining the $\displaystyle a_{n}$ is...

$\displaystyle \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1$ (2)

From (2) it follows that is...

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

I'm completely lost. We have a_n=1

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} = \frac{2}{3} < 1$ (3)

... so that the series (1) converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Hello,

How did we determine that...

$\displaystyle \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1$

I'm completely lost. We have $\displaystyle a_{n+1}=1$ and $\displaystyle a_n=0$.

Yet we have $\displaystyle \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1 }}{a_n}=\frac{2}{3}$

Wouldn't we have $\displaystyle \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1 }}{a_n} = \lim_{n\rightarrow\infty}\frac{1}{0}$ = undefined ?

And what does "DE" mean?

Sorry for all the questions, just trying to fully understand this.

8. Originally Posted by MechEng
Hello,

How did we determine that...

$\displaystyle \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1$

I'm completely lost. We have $\displaystyle a_{n+1}=1$ and $\displaystyle a_n=0$.

Yet we have $\displaystyle \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1 }}{a_n}=\frac{2}{3}$

Wouldn't we have $\displaystyle \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1 }}{a_n} = \lim_{n\rightarrow\infty}\frac{1}{0}$ = undefined ?

And what does "DE" mean?

Sorry for all the questions, just trying to fully understand this.

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)} = \sum_{n=1}^{\infty} a_{n}$

if u look at that all under sum as $\displaystyle a_n$

then your $\displaystyle a_{n+1}$ is:

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)(2n+1)}{1\ 4\ 7\ \dots \ (3n-2)(3n+1)} = \sum_{n=1}^{\infty} a_{n+1}$

"for $\displaystyle n+1$ is going to be $\displaystyle [2(n+1)-1]=(2n+1)$ and $\displaystyle [3(n+1)-2]=(3n+1)$ "

so when u put it in some of those criteria of convergence those a lot of members would be reduced...

like in D'alambert convergence criteria:

$\displaystyle \displaystyle q=\lim_{n\to\infty} \frac {a_{n+1}}{a_n}=\lim_{n\to\infty} \frac {\frac{1\ 3\ \ 5\ \dots \ (2n-1)(2n+1)}{1\ 4\ 7\ \dots \ (3n-2)(3n+1)}}{\frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)}}=\lim_{n\to\infty}\frac{2n+1}{3n+1}=\frac {2}{3}<1$

so it converges okay ?

Edit: Sorry i corrected that typo

9. Ah, yes... I follow that. Thank you!

Quick question...

$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{(2n-1)!}{(3n-2)!}$

Does make my statement any better?

10. Originally Posted by MechEng
Ah, yes... I follow that. Thank you!

Quick question...

$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{(2n-1)!}{(3n-2)!}$
Does make my statement any better?
no,

$\displaystyle (2n-1)!=(2n-1)(2n-2)(2n-3)... (2n-m)!$

so it's not

Edit: edited 2 times first time correct to incorrect and now back to correct (now i'm confused sorry )

maybe just maybe should be :

$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} =\sum_{n=1}^{\infty}\frac{\displaystyle \prod_{n=1}^{\infty}(2n-1)}{\displaystyle \prod_{n=1}^{\infty}(3n-2)}$

Edit: didn't know how to write product

11. Originally Posted by MechEng
Hello,

How did we determine that...

$\displaystyle \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1$

... and what does "DE" mean?...
Is...

$\displaystyle \displaystyle a_{n} = \frac{1\ 3\ 5\ \dots (2n-1)}{1\ 5\ 7\ \dots \ (3n-2)}$ (1)

... and...

$\displaystyle \displaystyle a_{n+1} = \frac{1\ 3\ 5\ \dots (2n-1)\ (2n+1)}{1\ 5\ 7\ \dots \ (3n-2)\ (3n+1)}$ (2)

... so that from (1) and (2) You obtain...

$\displaystyle \displaystyle \frac{a_{n+1}}{a_{n}} = \frac{2n+1}{3n+1}$ (3)

The term 'DE' means difference equation...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

12. Originally Posted by yeKciM
no,

$\displaystyle (2n-1)!=(2n-1)(2n-2)(2n-3)... (2n-m)!$

for n=1 it would be :

$\displaystyle 1\cdot 0\cdot ( -1)....$
But it's $\displaystyle (2n-1)!$

So...

$\displaystyle (2(1)-1)+(2(2)-1)+(2(3)-1)+(2(4)-1)+...+(2(n)-1)$

Or...

$\displaystyle 1+3+5+7+...+(2n-1)$

The n is what's changing.

Right?

13. Originally Posted by MechEng
But it's $\displaystyle (2n-1)!$

So...

$\displaystyle (2(1)-1)+(2(2)-1)+(2(3)-1)+(2(4)-1)+...+(2(n)-1)$

Or...

$\displaystyle 1+3+5+7+...+(2n-1)$

The n is what's changing.

Right?
hmmmm...
first there are no "+" there for Factorial

$\displaystyle 5!=5\cdot4\cdot3\cdot2\cdot1$

$\displaystyle 1\cdot3\cdot5\cdot. . .\cdot(2n-1)=\displaystyle \prod_{n=1} ^{\infty} (2n-1)$

but if yours is true then for let's say n=2 is going to be

$\displaystyle n=2$

$\displaystyle (2n-1)! = 3! = 3\cdot2\cdot1$

and that's not what u have on left side up there somewhere in starting sum

P.S. I corrected that post (#10)

14. Ah, yes...

Not sure how that happened.

15. Ok, I think I've gotten way too far off track here.

Starting with the original series:

$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5*...*(2 n-1)}{1*4*7*...*(3n-2)}$

What should my $\displaystyle \displaystyle a_n$ be equal to?

I understand all of the tests, but cannot seem to recognize the original series as something that I can work with.

I'm sure you've pointed out how to go from the original series to a more useable form, but I am having a hard time following you.

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