Converging Series Limit of pai: Proving Limit Not 1/2

[Abukadu]

hi :)
Quick questions:

How can I write a series converging into the limit of pai ?

Prove that the limit of the series zz (n+2) / (2n² +1) zz is not 1/2 when n->infinity (and bigger than 1)


I came to the equation:
zz |(-2n² + 2n +3) / (2n² +1) | >= ε zz

(*)And I made the left expression smaller by enlarging the denominator and diminution the numerator:

zz |(-2n² + 2n) / (2n² +n) | >= ε zz
which is
zz |(-2n + 2) / (2n +1) | >= ε zz
and again.. (*)
zz |(-2n) / (2n +n) | >= ε zz
which is
zz |-2/3 | >= ε zz

and then I can choose ε = 2/3 ?
something dosent seem right ..

 

[HallsofIvy]

Are those "zz"s suppose to delimit mathematical expressions? Please don't do that- especially not without explanation!

hi :)
Quick questions:

How can I write a series converging into the limit of pai ?

Prove that the limit of the series zz (n+2) / (2n² +1) zz is not 1/2 when n->infinity (and bigger than 1)

The simplest thing to do is to divide both numerator and denominator by n² so that you get ((1/n)+ (2/n²))/(2+ (1/n)). Now you should be able to see that, as n goes to infinity, those fractions go to 0.

I came to the equation:
zz |(-2n² + 2n +3) / (2n² +1) | >= ε zz

(*)And I made the left expression smaller by enlarging the denominator and diminution the numerator:

zz |(-2n² + 2n) / (2n² +n) | >= ε zz
which is
zz |(-2n + 2) / (2n +1) | >= ε zz
and again.. (*)
zz |(-2n) / (2n +n) | >= ε zz
which is
zz |-2/3 | >= ε zz

and then I can choose ε = 2/3 ?
something dosent seem right ..