help on sequences and series. .

[Lost souls]

i hav to do a convergence test on the following series but m having trouble finding the n^th term. .cn sum1 take a look pls. .?
pic attached. .i m having difficulty starting so no work to show. .

 

[mmm4444bot]

cn sum1 take a look

i m having difficulty starting so no work to show

I see sum1, but where is sum2 ? :cool:

Seriously, please drop the text-speak. (There is no place in mathematics for sloppy communication; it just wastes time.)

You do not need to show work, to make a start. You may explain what you've thought about.

Are you having difficulty writing the numerators in terms of n?

Have you looked for a pattern in the denominators?

Here is an expression for the product of the first k odd Natural numbers:

1*3*5*…*k = (2k)!/(2^k*k!)

 

[soroban]

Hello, Lost souls!

\(\displaystyle \text{Find the }n^{th}\text{ term:}\)

. . . . \(\displaystyle 1 + \dfrac{1^2\cdot2^2}{1\cdot3\cdot5} + \dfrac{1^2\cdot 2^2\cdot3^2}{1\cdot3\cdot5\cdot7\cdot9} + \cdots \)

Let's examine the third term \(\displaystyle (n=3).\)


The numerator is: .\(\displaystyle N_3 \;=\;1^2\cdot2^2\cdot3^2 \;=\;(1\cdot2\cdot3)^2 \;=\;(3!)^2\)

. . In general, the numerator is: .\(\displaystyle N \;=\;(n!)^2\)


The denominator is: .\(\displaystyle D_3 \;=\;1\cdot3\cdot5\cdot7\cdot9\)

This is the product of odd integers from 1 to 9 \(\displaystyle (4n-3).\)

We start with \(\displaystyle 9! \:=\:1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8 \cdot9\)

And we cancel the even integers between 1 and 9: .\(\displaystyle 2\cdot4\cdot6\cdot8\)

. . The denominator looks like this: .\(\displaystyle D_3 \;=\;\dfrac{1\cdot2\cdot3 \cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot4\cdot6\cdot8}\)

Note that: .\(\displaystyle 2\cdot4\cdot6\cdot8 \;=\;(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4) \;=\;2^4(4!)\)

. . Hence, the denominator is: .\(\displaystyle D_3 \;=\;\dfrac{9!}{2^4(4!)} \)


Now we must generalize this result.

Its numerator is \(\displaystyle (4n-3)!\)

Its denominator has half that many terms: .\(\displaystyle \frac{(4n-3)-1}{2} \,=\,2n-2\)

and is the product of \(\displaystyle 2n-2\) consecutive even numbers: .\(\displaystyle 2^{2n-2}(2n-2)!\)

. . Hence, the denominator is: .\(\displaystyle D \;=\;\dfrac{(4n-3)!}{2^{2n-2}(2n-2)!}\)


Therefore, the n^th term is: .\(\displaystyle a_n \;=\; \dfrac{(n!)^2} {\frac{(4n-3)!}{2^{2n-2}(2n-2)!}} \;=\;\dfrac{2^{2n-2}(n!)^2(2n-2)!}{(4n-3)!} \)
I need a nap!

 

[Lost souls]

thanx soroban. . .that was a pretty detailed and well explained reply. . i'll hav a hard time applying the tests on this expression,methinks. .

ps: mmmm444bot, i'll remember nxt time. .this habit has gone out of hands. .