Question

Alternating Series Test help

Answer 1

Recall that the Alternating Series Test has three conditions associated with it:
1. The series must alternate.
2. The terms must decrease (in absolute value) for large n.
3. The nth term must go to 0.

Can you even have a series where the first two conditions hold and the third doesn't?
Create a convergent series that satisfies conditions 1 and 3, but not 2.
Create a divergent series that satisfies conditions 1 and 3, but not 2

Answer 2

This is interesting... you'll have my answer by your return, @Best_Mathematician
:D

Answer 3

For a series that alternates and decreases, but does not go to 0, just find a series that decreases but not to 0 e.g. \(\frac{n+1}n\). Throw in a factor of \((-1)^n\) to make it alternating:$$a_n=(-1)^n\frac{n+1}n$$

Answer 4

@terenzreignz can you solve it

Answer 5

what about last two questions

Answer 6

ok I lied so (n+1)/n is not decreasing. it simplifies to just over one for every n. ok so lets look at a divergent series like sin(x) since the question does not specify an infinite series let us say that n approaches pi/2

Answer 7

does this satisfy converging to zero when n=pi/2

Answer 8

i guess

Answer 9

but it both alternates and decreases if you add (-1)^n right?

Answer 10

ya

Answer 11

and best... what is the sin (pi/2)???

Answer 12

1

Answer 13

crap I messed up disregard everything I just posted please

Answer 14

ok lol

Answer 15

its increasing on an absolute value hmm do from n=1 to n=-pi/2 then it works

Answer 16

but can you have a negative bound on a series... of that I am not certain

Answer 17

\[a_n=\left\{\begin{matrix}\frac{1}{n^2} & n\text{ even} \\ -\frac{1}{n^3} & n\text{ odd}\end{matrix}\right.\]

Answer 18

and

\[a_n=\left\{\begin{matrix}\frac{1}{n} & n\text{ even} \\ -\frac{1}{n^2} & n\text{ odd}\end{matrix}\right.\]

Answer 19

i think you posted same thing twice

Answer 20

my bad...nm

Answer 21

both are convergent series right

Answer 22

no

Answer 23

why dont they approach to zero

Answer 24

both sequences go to zero as n goes to infinity...but the 1st series converges and the second diverges

Answer 25

ohk

Answer 26

so the first is answer of second question and second is answer of third?

Answer 27

yes

Answer 28

arent they all decresing too @Zarkon

Answer 29

no

Answer 30

how

Answer 31

plug in some values for n...that should convince you

Answer 32

not really the larger n gets the smaller a p series hence a decreasing dunction

Answer 33

for \[a_n=\left\{\begin{matrix}\frac{1}{n} & n\text{ even} \\ -\frac{1}{n^2} & n\text{ odd}\end{matrix}\right.\]

\[|a_1|=1\]
\[|a_2|=1/2\]
\[|a_3|=1/9\]
\[|a_4|=1/4\]
\[|a_5|=1/25\]
\[|a_6|=1/6\]
\[|a_7|=1/49\]
\[|a_8|=1/8\]

doesn't look like it is a decreasing sequence to me

Answer 34

I may be putting my foot in my mouth with this- but that is not a single alternating series. It defines even and odd partial sums. Please educate me if I am wrong(not sarcasm), but that would not work for this particular question because I do not thik that qualifies as a series.(I can definitely be wrong and probably am, can you direct me to some literature on the subject if this is the case)

Answer 35

"...not a single alternating series."

if it is not a 'single AS' then what is the other part?

it looks to me that for each \(n\in\mathbb{N}\) you get one value \(a_n\)

Answer 36

but it is two different series for even and odd values, I'm not sure if that is allowed with series like functions

Answer 37

\[a_n=\frac{1}{2n}-\frac{1}{2n^2}+(-1)^n\left[\frac{1}{2n}+\frac{1}{2n^2}\right]\]

how about that

Answer 38

\[a_n=\frac{1}{2n}-\frac{1}{2n^2}+(-1)^n\left[\frac{1}{2n}+\frac{1}{2n^2}\right]\]

if \(n\) is even then
\[a_n=\frac{1}{n}\]

if \(n\) is odd then
\[a_n=-\frac{1}{n^2}\]

Answer 39

looks like one thing to me

Answer 40

@FibonacciChick666 #1 (n+1)/n = 1 + 1/n is definitely decreasing as n -> infty... besides, convergence makes no sense when dealing with finite series.

Answer 41

ahhh, that makes more sense ok. but I still disagree with the allowance of even and odd in the definition. i would simplify this to \[\frac{(1+(-1)^n)}{2n}+\frac{((-1)^n-1)}{2n^2}\] which for n=1 is -1/n^2

Answer 42

@oldrin.bataku yea I retracted, I realized that my adding sucks

Answer 43

ok @Zarkon I disagree with your original notation but now understand your meaning.

Answer 44

when you put \((-1)^n\) in the definition of a sequence you are essentially defining it for odds and evens.

Answer 45

@Zarkon right but a piecewise definition like that is not closed-form, which might be a problem.

Answer 46

um...yes it is

Answer 47

but prior to writing it out in terms it was not accurate for this answer I don't think

Answer 48

http://en.wikipedia.org/wiki/Closed-form_expression

Answer 49

I know very well what a closed-form expression is, @Zarkon, but that article does not agree with you. Additionally, a quick Google search yields:
http://onlinelibrary.wiley.com/doi/10.1002/cta.376/abstract

Answer 50

then you can't read

Answer 51

but what you wrote was not a series that satisfied the original question the first time, the expansion where you can see the terms go to zero yeilding that simplified expression is better for a this question. it's like not showing all of your work and just writing an answer in the end it is incorrect and no credit

Answer 52

though that expression is what the series simplifies to, it is not the series

Answer 53

what i wrote works...and is a solution to this problem

Answer 54

Still nothing agrees with what you're saying. A piecewise-defined function is not presented using a single closed-form analytic expression. Just because it may be broken into pieces which are indeed closed-form analytic expressions doesn't mean the whole is. I'm not arguing that what you wrote is wrong, but I believe @FibonacciChick666 is entirely correct in pointing out OP is more than likely expected to present the solution using a single closed-form analytic expression.

Answer 55

Thank you @oldrin.bataku you explained that much more effectively

Answer 56

ok guys dont take it so high. this is just high school level calculus which i need help with

Answer 57

I don't know why you responded so aggressively.

Answer 58

Does it say you cant write it as a piece wise function in the directions...it jusrt said give an example...and I did

Answer 59

Nobody is doubting that you gave a working solution. The thing is, since this is high-school calculus, OP is more than likely going to want a single closed-from analytic expression. Again, you did indeed give a working solution, but @FibonacciChick666 presented it in a form that @Best_Mathematician is more likely to be able to use.

Answer 60

where is "FibonacciChick666 "'s solution ?

Answer 61

ya...i really can't find it....there are lot of comments and stuff

Answer 62

the expanded and simplified version of your piece-wise series is what he was referring to

Answer 63

I don't recall ever calling it @FibonacciChick666's solution, it's your's @Zarkon... it was just rewritten using \((-1)^n\) tricks to condense it into a single expression.

Answer 64

I'm refering to this...

"...but @FibonacciChick666 presented it in a form that @Best_Mathematician is more likely to be able to use."

Answer 65

please help me out guys....no time for this stuff

Answer 66

Yes, exactly -- he presented your solution in a different more likely usable way.

Answer 67

she is a girl

Answer 68

so do i got answers?