# infinite sequences and series

• Nov 28th 2009, 08:12 PM
noscbs
infinite sequences and series
T= from n=0 to infinity of k^(2n) x ((1x3x5...(2n-1))/(2x4x6x...(2n)))

or
T=[1+(1^2/2^2) k^2 + (1^2x3^2/2^2x4^2) k^4 ...]

notice that all the terms in the series after the first one have coefficients that are at most 1/4. Use this fact to compare this series with a geometric series and show that

(1+k^2/4) < T < (4-3k^2)/(4-4k^2)
• Nov 28th 2009, 08:33 PM
RockHard
Quote:

Originally Posted by noscbs
T= from n=0 to infinity of k^(2n) x ((1x3x5...(2n-1))/(2x4x6x...(2n)))

or
T=[1+(1^2/2^2) k^2 + (1^2x3^2/2^2x4^2) k^4 ...]

notice that all the terms in the series after the first one have coefficients that are at most 1/4. Use this fact to compare this series with a geometric series and show that

(1+k^2/4) < T < (4-3k^2)/(4-4k^2)

$\displaystyle \sum_{n=0}^{\infty}k^{2n}\cdot\frac{2n-1}{2n}$

Is this your series, correct? Can this be a power series? I am learning the same subject as well, and just as well curious to get all facts before I attempt this problem

Also can someone rectify for me since this function would be undefined at n = 0 if the current form I wrote above, is correct before doing some simplification, would you note want to do an index shift and start the index and n = 1?
• Nov 28th 2009, 08:55 PM
simplependulum

$\displaystyle \sum_{n=0}^{\infty} k^{2n} \frac{ 1 \cdot 3 \cdot 5 ... \cdot (2n-1)}{2 \cdot 4 \cdot 6 ... \cdot (2n) }$

If we multiply the numerator and the denominator by $\displaystyle 2 \cdot 4 \cdot ... \cdot (2n)$

( the purpose is to fill up the space between odd no. )

the numerator becomes

$\displaystyle k^{2n} 1 \cdot 2 \cdot 3 \cdot 4 ... \cdot (2n-1) \cdot (2n) = (2n)!$

the denomerator becomes

$\displaystyle \left (2 \cdot 4 \cdot 6 ... \cdot (2n) \right )^2$

If we the take a factor $\displaystyle 2$ out from each even no.
and since there are n terms , it is equivalent to

$\displaystyle = \left( 2^n 1 \cdot 2 \cdot 3 \cdot ... \cdot n \right )^2$

$\displaystyle = 4^n (n!)^2$

therefore the series is actually

$\displaystyle \sum_{n=0}^{\infty} \left ( \frac{k}{2} \right )^{2n} \frac{(2n)! }{ (n!)^2 }$

$\displaystyle = \sum_{n=0}^{\infty} \left ( \frac{k}{2} \right )^{2n} \binom{2n}{n}$
• Nov 28th 2009, 09:20 PM
RockHard
Amazing, This is why calculus make me mad at times, how do you know when to do such things or if it is proper/correct to do that?
I am having trouble trying to write out the work for the last bit where you begin the factoring
• Nov 28th 2009, 10:13 PM
simplependulum
Quote:

Originally Posted by RockHard
Amazing, This is why calculus make me mad at times, how do you know when to do such things or if it is proper/correct to do that?
I am having trouble trying to write out the work for the last bit where you begin the factoring

But my expression of the series is not useful to solve the problem ...

In fact , it is the Taylor's expression for the function $\displaystyle \frac{1}{\sqrt{1 - k^2}}$

some textbooks may show how to reach the result (Happy)
• Nov 28th 2009, 10:54 PM
RockHard
Thanks for the input because those factorial are tricky to me sometimes.
• Nov 29th 2009, 05:57 PM
noscbs
Yes rockhard I completely agree that these factorials can be tricky in this problem ... however, the first part of the inequality seems pretty easy, T will always be larger, because you are adding a positive value to something of an equal value. It is the second part, proving T is less than that expression, that I just cannot get.
• Nov 29th 2009, 06:45 PM
RockHard
Agreed. Also getting rid of factorials when simplifying for me gets confusing alot, hence why I have trouble grasping SP last reply the second part, more so