Question

Series !

Answer 1

whats the question ^_^

Answer 2

What is the question?

Answer 3

What is the...
Oh never mind XD

Answer 4

\[\sum_{n=1}^{\infty}\frac{ 1 }{ (2n-1)(2n+1)(2n+3) }\]

Answer 5

O_O ^ idk what that even means so ya...... maybe the master at mathematics can help you @hartnn take control dont run away bruh xD

Answer 6

Oh, the humanity :D

Answer 7

^ LOL XD

Answer 8

Hold on, I think I can help.

Answer 9

Holding on... ;)

Answer 10

^ smart girl so... ya im out ill go check on other questions have fun

Answer 11

@ganeshie8 @waterineyes @radar @TQKMB @unicornsandmarshmellows @dumbcow @hartnn @madrockz

Answer 12

Is it asking you to evaluate or evaluate using summation formulas?

Answer 13

@Rachella ever done calculus series ?

Answer 14

I have :D

Answer 15

@SolomonZelman

Answer 16

I have. I'm currently taking Pre-Calculus.

Answer 17

in pre-calculus you don't do series

Answer 18

I know. I'm saying, I've taken Series in the past.

Answer 19

the only thing you can ask for a series problem is study the nature

Answer 20

Guys, let's not fight D:

Let Miss @Rachella do her thing XD

Answer 21

that's not that kind of series
Those made by you were very particular examples easy to summate

Answer 22

see if it telescopes

Answer 23

the general strategy is to decompose it into partial fractions and try to cleverly split it into two telescoping sums

Answer 24

\[\sum_{n=1}^{\infty} \frac{ 1 }{ (2n-1)(2n+1)(2n+3)} = Converges\]

Answer 25

Without justifying?

Answer 26

\[\begin{align}
&\sum_{n=1}^{\infty}\frac{ 1 }{ (2n-1)(2n+1)(2n+3) } = \frac{1}{8}\sum_{n=1}^{\infty} \left(\dfrac{1}{2n-1} -\dfrac{2}{2n+1}+\dfrac{1}{2n+3}\right)\\~\\

\end{align}\]

Answer 27

split the middle term into two and you will get two telescoping sums

Answer 28

... I knew that... haha

peace, @ganeshie8
Kudos for having the patience to do those partial fractions ^^

Answer 29

wolfram did that, not me lol

Answer 30

*shrug* Kudos to wolfram, then, but I can't give wolfram any medals XD

Answer 31

@SithsAndGiggles i was just doubting if that splitting of series into two different telescoping sums is technically correct.. it seems to work for this particular case but im still skeptical as we can evaluate that series to any other number by splitting it in a different way :O

Answer 32

I remember reading something like a converging series can be cooked up to evaluate to any real number by rearranging... pulling up my analysis notes...

Answer 33

That'd be interesting to see for this series... We know it converges by a comparison test to \(\dfrac{1}{n^3}\), so there should only be one value to which it converges (limits are unique, etc). I think this approach is solid.

Answer 34

I mean rearrangement like this :

we know below alternating harmonic series converges as it is written
1 -1/2 + 1/3 -1/4 + ...

Answer 35

but if we rearrange it like

1 +1/3 + 1/5 - 1/6 + 1/9 + 1/11 - 1/12 + ....

by correcting the error continuously we can make it evaluate to the number : (1 +1/3 + 1/5 )

not really sure if i understood this thing correctly

Answer 36

Yeah I think you have it right: http://en.wikipedia.org/wiki/Alternating_series#Rearrangements

However, this only seems to be the case for alternating series, or at least those which converge conditionally. We know for a fact that the given series converges absolutely.

Answer 37

Oh right! the terms are positive in the given series so there is no question of correcting the errir yeah. didn't think of that thanks :)

Answer 38

np!