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Interval of Convergence for acrtan(x) series

So I derived the general formula for each term in the power series of arctan(x) :



n=0n=(1)nx2n+12n+1\sum_{n=0}^{n=\infty}\frac{(-1)^n x^{2n+1}}{2n+1}



I then used the ratio test and simplified to get :

limn2n+12n+3x2\lim_{n \to\infty } \left | \frac{2n+1}{2n+3} \right |\left | x^2 \right |

Giving just lx^2l which means the series only converges for :

-1<x^2 <1

At this point I'm stuck on what to do. How do I proceed ? I cant simply take a square root of each side....
(edited 11 years ago)
Reply 1
Original post by Ari Ben Canaan
So I derived the general formula for each term in the power series of arctan(x) :



n=0n=(1)nx2n+12n+1\sum_{n=0}^{n=\infty}\frac{(-1)^n x^{2n+1}}{2n+1}



I then used the ratio test and simplified to get :

limn2n+12n+3x2\lim_{n \to\infty } \left | \frac{2n+1}{2n+3} \right |\left | x^2 \right |

Giving just lx^2l which means the series only converges for :

-1<x^2 <1

At this point I'm stuck on what to do. How do I proceed ? I cant simply take a square root of each side....


There is not x^2 at the limit when you calculate the radius.
So your radius is 1.
The series is given around 0
the range of convergence |x-0|<1
Reply 2
Original post by Ari Ben Canaan
Giving just lx^2l which means the series only converges for :

-1<x^2 <1

At this point I'm stuck on what to do. How do I proceed ? I cant simply take a square root of each side....
Why not?

Spoiler

Original post by DFranklin
Why not?

Spoiler



Sorry, I meant to say taking the square root would mean getting i on the LHS.

Hmmm, so essentially the interval of convergence is just 0<x<1 ?
Original post by Ari Ben Canaan
Hmmm, so essentially the interval of convergence is just 0<x<1 ?


You're forgetting the "other half"
Original post by Lord of the Flies
You're forgetting the "other half"


I don't quite follow...
Original post by ztibor
There is not x^2 at the limit when you calculate the radius.
So your radius is 1.
The series is given around 0
the range of convergence |x-0|<1


What do you mean there is no x^2 ? Of course there is....
Original post by Ari Ben Canaan
I don't quite follow...


I was hinting at negative values of x (also, don't forget to check endpoints!).
Original post by Lord of the Flies
I was hinting at negative values of x (also, don't forget to check endpoints!).


-1<x<1 ?

Mu question as to how to deal with the x^2 still hasn't been answered or at least I haven't understood it yet. Could you please just spell out for me ?
Original post by Ari Ben Canaan
-1<x<1 ?

Mu question as to how to deal with the x^2 still hasn't been answered or at least I haven't understood it yet. Could you please just spell out for me ?


There's nothing to deal with really: 0x2<11<x<10\leq x^2<1\Rightarrow -1<x<1

Since the ratio test is inconclusive when x=±1x=\pm 1 you need to check if the series converges at these values (it does).

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