Another series I'm testing for convergence....

**Malaclypse** · April 21st 2011, 04:11 PM

Sorry to bother with another similar question to the one I was asking earlier, but I have one more series that I'm stuck on...

$\sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n - 1)}}{{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)}}}$

I'm pretty sure I did the cancellation right on this one:

$\rho = \mathop {\lim }\limits_{n \to \infty } \frac{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n + 1)}}{{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n + 2)}} \cdot \frac{{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)}}{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n - 1)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{(2n + 1)}}{{(2n + 2)}} = 1$

How do you tackle this kind of series when that test fails? I lean towards the comparison test, but I'm having trouble thinking of a comparable series that would appropriately converge or diverge to provide me with meaningful information.

**chisigma** · April 21st 2011, 07:06 PM

Originally Posted by Malaclypse

Sorry to bother with another similar question to the one I was asking earlier, but I have one more series that I'm stuck on...

$\sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n - 1)}}{{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)}}}$

I'm pretty sure I did the cancellation right on this one:

$\rho = \mathop {\lim }\limits_{n \to \infty } \frac{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n + 1)}}{{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n + 2)}} \cdot \frac{{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)}}{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n - 1)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{(2n + 1)}}{{(2n + 2)}} = 1$

How do you tackle this kind of series when that test fails? I lean towards the comparison test, but I'm having trouble thinking of a comparable series that would appropriately converge or diverge to provide me with meaningful information.

Remembering the 'binomial expansion'...

(1)

... and setting in (1) x=1 You find that the series diverges...

Kind regards

$\chi$ $\sigma$

**Malaclypse** · April 21st 2011, 08:29 PM

Thank you very much. This is an area we have not covered yet, and I will need to study it.

**chisigma** · April 21st 2011, 10:19 PM

Alternatively You can use the [not very 'popular'...] 'Raabe test' according to which, given a positive terms series...

... if for n 'large enough' ...

(1)

... the series converges and if...

(2)

... the series diverges. In your case is...

(3)

... so that the series diverges...

Kind regards

$\chi$ $\sigma$

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