Question

Determine whether the series is absolutely convergent, conditionally convergent, or divergent

Answer 1

\[\frac{ 3 }{ 4 }+\frac{ 3*7 }{ 4*7}+\frac{ 3*7*11 }{ 4*7*10 }\]

Answer 2

...

Answer 3

the next one in line would be \[\frac{ 3*7*11*15 }{ 4*7*10*13 }\]

Answer 4

i dont have a clue where to start here

Answer 5

For what it's worth, you have the following closed form for the series:
\[{\Large\sum_{n=1}^\infty}\frac{\displaystyle\prod_{k=0}^{n-1}(4k+3)}{\displaystyle\prod_{k=1}^n(3k+1)}\]

Answer 6

Hmm, try the ratio test?

Answer 7

thank you for the simplification there, although even simplified, it is confusing

Answer 8

"Closed form" doesn't always mean "simplified" :P

Answer 9

ok, i will have to look up closed form real quick

Answer 10

I feel like I seen a trick to this one but I can't remember off the top of my head.

Answer 11

freckles!

Answer 12

Like on Openstudy recently

Answer 13

i do believe it is the ratio test... it is soooo much tho

Answer 14

wow, this one MUST be hard, if you two dont know :(

Answer 15

Via the ratio test,
\[\lim_{n\to\infty}\left|\frac{\displaystyle\prod_{k=0}^{n}(4k+3)}{\displaystyle\prod_{k=1}^{n+1}(3k+1)}\times\frac{\displaystyle\prod_{k=1}^n(3k+1)}{\displaystyle\prod_{k=0}^{n-1}(4k+3)}\right|=\lim_{n\to\infty}\left|\frac{4n+3}{3(n+1)+1}\right|\]

Answer 16

ok,.... so, I need to learn how to do the closed form to find the initial values you got. then , is this \[\frac{ a_n+1 }{ a_n }\]?

Answer 17

it looks like just a_n flipped over, thats y i was wondering

Answer 18

You don't *need* the closed product formula for this one. The ratio test alone will show that a lot of lagging terms will cancel. And yes, the limit expression is \(\dfrac{a_{n+1}}{a_n}\).

Answer 19

4/3

Answer 20

Right, which means this series is...

Answer 21

\[\frac{ 4 }{ 3 }>1\].... divergent?

Answer 22

Correct

Answer 23

omg... idk how i would do this problem without learning what you did!

Answer 24

and i couldn't find any tricks
i think it was just my mind playing tricks on me that there was some other trick or whatever
anyways @SithsAndGiggles thingy worked out gorgeously

Answer 25

for real! I thank you kindly!!!

Answer 26

it is funny how it ended up being \[\frac{ 4n+3 }{ 3n+4 }\]

Answer 27

Could either of you tell me if this could be considered a "taylor" series or something?

Answer 28

i have several of these types of questions on my homework and I cant figure out how to do them. I don't even know what to call them. If you could lead me to something I could study myself, i would forever be grateful to you

Answer 29

@SithsAndGiggles and @freckles

Answer 30

You can find plenty of info and examples here: http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx