Question

Question on limit and wolfram. I entered the limit (-1)^n, as n->infinity. I know this limit diverges, but wolfram says its e^(2Pi * i ) from 0 to pi
any ideas why wolfram says this
http://www.wolframalpha.com/input/?i=limit+%28-1%29^n%2C+n+%3D+oo
of course e^(2Pi * 0 ) = 1 , e^(2pi * pi/2) = -1
but why does it say 0 to pi . Surely not every number between 0 and pi is a solution

Answer 1

on maple I entered the same expression and I got this output
-1-I .. 1+I

Answer 2

theres something going on inside the program, it converts such limits to complex numbers , maybe a gamma function

Answer 3

there are two cases x is odd or x is even

Answer 4

right, the sequence is 1,-1,1,-1 ...

Answer 5

i was just wondering why wolfram and maple give such strange answers

Answer 6

or -1,1,-1,.. depending on how you start n

Answer 7

do u think the limit exists

Answer 8

no it definitely diverges

Answer 9

there should be something called divergence test criteria for this

Answer 10

thats for series, this is a limit of sequence

Answer 11

http://www.wolframalpha.com/input/?i=sum+x-%3E%5Cinfty+%28-1%29%5E%7Bn%7D

Answer 12

my question was a limit, not a series

Answer 13

i think you have a typo there,

Answer 14

you wrote x-> infinity, it should be n
here i fixed it
http://www.wolframalpha.com/input/?i=sum+n-%3E \infty+%28-1%29^{n}

Answer 15

oh right

Answer 16

thats annoying when you paste a link with a \ in it, it won't show up in blue

Answer 17

here
http://www.wolframalpha.com/input/?i=Sum+%28-1%29^n+%2C+n%3D1+..+infinity

Answer 18

http://en.wikipedia.org/wiki/Term_test

see this if the limit doesnt exist then it diverges

Answer 19

so u have to just prove it doesnt exists

Answer 20

correct.
limit an = 0 as n->oo is a necessary condition for series an to converge (but not a sufficient condition)

Answer 21

isn't that just the bounds of the limit?

Answer 22

if series an converges, then lim an = 0 as n->oo.

The contrapositive of this is the 'divergence test'

if lim an =/= 0 , then series an does not converge

Answer 23

inky, the bounds are n=1 or some positive integer to n= infinity

Answer 24

@perl I mean it's the range of possible results for n... for instance (-1)^.5=i

Answer 25

@mathmath333 do you see why lim a_n = 0 is not a sufficient condition for the series a_n to converge ?

is it true that If lim a_n=0 , must the series an converge?

Answer 26

inky are you answering the question in the original post above

Answer 27

yeah derp

Answer 28

n is positive integers

Answer 29

you can start at some positive integer

Answer 30

if you want fractions, then you can use
(-1)^x , but thats a whole other expression

Answer 31

but maybe thats what wolfram is doing

Answer 32

the way I see it wolfram has simply determined that your function is periodic with a domain in the hyperreals and a range in the complex plane

Answer 33

http://www.symbolab.com/solver/calculus-calculator/%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft(-1%5Cright)%5E%7Bn%7D/?origin=button

Answer 34

because if you consider (-1)^x for any real x I would assume you get the argand diagram of e^(i x)

Answer 35

so instead of saying the limit does not converge to one value wolfram simply tells you that it oscillates between certain values

Answer 36

ok so wolfram is looking at all x real numbers?
i thought it assumes n is discrete integers

Answer 37

-1 = e^ ( Pi * i )

so (-1)^x = e ^ ( Pi * i * x )

Answer 38

or

(-1)^n = e^(Pi * i * n )

Answer 39

I guess

Answer 40

what do you mean by 'hyperreals' ?

"the way I see it wolfram has simply determined that your function is periodic with a domain in the `hyperreals` and a range in the complex plane"

Answer 41

did you say hyperreals because you are including infinity in your domain?

Answer 42

@inkyvoyd

Answer 43

wolfram value means its div
when power is 0 then its 1 otherwise its -1

Answer 44

@Marki sorry i didnt understand you
when power is 0 , then its 1. if power is not 0 , then its -1?
are you talking about
e^(2pi * i )

Answer 45

yes
note its from 0 to pi

Answer 46

@Marki you get some strange answers if you plug in all reals into
e^(2*i*x) for x between 0 and Pi
the range are points on the unit circle

for example , when x= 0.5 the output is .5403 + .8414i
when x = 2 you have -.6536 - .7568 i
So i'm not sure why wolfram would do that.