weird series: Z_3 = 1/2, Z_n= Z_{n+1}/[cos(pi/n)]. It's a bizarre suite decreasing to 0 without ever going.

extrazlove

New member
Joined
Feb 15, 2019
Messages
2
hello

Here is a Zn Suite



Z3 = 1/2
Zn= Zn+1 / \cos (pi / n)


It's a bizarre suite decreasing to 0 without ever going.




I will define the n with which I work in my suite so as not to fall into absurdities like 1/0 which gives false calculations.
find that


n = pi / arccos (Zn+1 / Zn)


Note that Zn #0 and Zn+1/ Zn #1
so the Zn+1 limit is non-zero.
And since I have a decreasing sequence, Zn+1 / Zn < 1, so the series Zn is convergent according to the d'Alembert criterion.

Is there an error in this reasoning?

Zn is a positive term and is decreasing and minus 0 tends to 0 minus bound 0, which is impossible because Zn #0?
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,795
So \(\displaystyle Z_3= 1/2\) and \(\displaystyle Z_n= \frac{Z_{n+1}}{cos(\pi/n)}\)? That second equation is oddly written. I would write it as \(\displaystyle z_{n+ 1}= z_n cos(pi/n)\).

You say "Zn is a positive term and is decreasing and minus 0 tends to 0 minus bound 0, which is impossible because Zn #0?"
\(\displaystyle Z_n\) is never 0 but that surely does not mean that the limit cannot be 0! The very simple sequence \(\displaystyle \{\frac{1}{n}\}\) has that same property: no term is ever 0 but the limit is 0.
 
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