I Finding a limit involving Chebyshev polynomials

Unconscious
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How could I show that this limit:

##\lim_{N\to\infty}\frac{\sum_{p=1}^N T_{4N} \left(u_0(N)\cdot \cos\frac{p\pi}{2N+1}\right)}{N}##

is equal to 0?

In the expression above ##T_{4N}## is the Chebyshev polynomials of order ##4N##, ##u_0(N)\geq 1## is a number such that ##T_{4N}(u_0)=b##, with ##b\geq 1## fixed.

I tried to write ##T_{4N}## in its polynomial form, and to expand in series the terms ##\cos^k##, trying to reach a geometric series that would simplify everything to me in a chain, but still remains an abomination.
 
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Chebyshev polynomials are defined over [−1,1], but you have ##T_{4N}(u_0)=b## with ##b \ge 1##. Are you extending the domain to ## [-b,b]##? NO! You can't mangle up the argument of the Chebyshev polynomials to fit your pathological conjecture.
 
Keep calm.
No pathological conjectures, no conspiracy theories against Chebyshev's polynomials.

Fred Wright said:
Chebyshev polynomials are defined over [−1,1]
False.
Read here: https://en.wikipedia.org/wiki/Chebyshev_polynomials#Explicit_expressions .

For other references on this limit, read: R. S. Elliott, Antenna Theory and Design, Linear Array Synthesis.
 
Unconscious said:
Keep calm.
No pathological conjectures, no conspiracy theories against Chebyshev's polynomials.False.
Read here: https://en.wikipedia.org/wiki/Chebyshev_polynomials#Explicit_expressions .

For other references on this limit, read: R. S. Elliott, Antenna Theory and Design, Linear Array Synthesis.
Ok, my bad. Have you tried evaluating the sum with ##N=1##?
 

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