Series Convergence question dealing with 1*3*5* and n!

**ihatemath** · November 2nd 2010, 08:42 PM

Determine whether series converges or diverges.

(1*3*5***(2n-1))/n!

I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.

**bkarpuz** · November 2nd 2010, 10:46 PM

Originally Posted by ihatemath

Determine whether series converges or diverges.

(1*3*5***(2n-1))/n!

I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what $1*3*5***$ means, and I do not know how to approach or solve the problem.

$1*3*5***(2n-1)$ means multiplication of first $n$ odd numbers from $1$ to $(2n-1)$ .
In general it is shown as
$\prod\limits_{k=1}^{n}(2k-1)$ .
I guess that your series is
$\sum\limits_{n=1}^{\infty}\dfrac{1\times3\times\cd ots\times(2n-1)}{n!}$ ,
which can be also represented as
$\sum\limits_{n=1}^{\infty}\dfrac{1}{n!}\prod\limit s_{k=1}^{n}(2k-1)$ .
Your general term for this series is
$a_{n}:=\dfrac{1\times3\times\cdots\times(2n-1)}{n!}$ .
So that, we have
$\dfrac{a_{n+1}}{a_{n}}=\dfrac{\dfrac{1\times3\time s\cdots\times\big(2(n+1)-1\big)}{(n+1)!}}{\dfrac{1\times3\times\cdots\times (2n-1)}{n!}}$
......... $=\dfrac{\dfrac{1\times3\times\cdots\times(2n+1)}{( n+1)!}}{\dfrac{1\times3\times\cdots\times(2n-1)}{n!}}$
......... $=\dfrac{\dfrac{1\times3\times\cdots\times(2n-1)\times(2n+1)}{n!(n+1)}}{\dfrac{1\times3\times\cd ots\times(2n-1)}{n!}}$
......... $=\dfrac{2n+1}{n}$
By using the ration test, we get
$\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=\li m\limits_{n\to\infty}\dfrac{2n+1}{n}$
.................. $=2>1$ ,
which shows that your series is divergent.

**chisigma** · November 3rd 2010, 06:00 AM

Originally Posted by ihatemath

Determine whether series converges or diverges.

(1*3*5***(2n-1))/n!

I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.

The general term of the series is...

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

... and because is $\displaystyle \lim_{n \rightarrow \infty} a_{n} = \infty$ the series [strongly] diverges...

Kind regards

$\chi$ $\sigma$

**tonio** · November 3rd 2010, 06:55 AM

Originally Posted by ihatemath

Determine whether series converges or diverges.

(1*3*5***(2n-1))/n!

I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.

$\frac{1\cdot 3\cdot\ldots\cdot (2n-1)}{n!}=\frac{(2n)!}{2^nn!}$ .

Now use the quotient test to prove at once that the series diverges.

Tonio

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