How could I approach \lim_{n \to \infty} (\frac{1+\co

Mabel Breault

Mabel Breault

Answered question

2022-01-02

How could I approach
limn(1+cos(12n)2)n
I tried several things to evaluate it, namely looking at it as cos(12n+1)2n instead or as exp(2nln(cos(12n+1)) and then trying to show that the limit of nln(cos(12n+1)) is 0 (for example using LHopitals rule), but I havent

Answer & Explanation

Jim Hunt

Jim Hunt

Beginner2022-01-03Added 45 answers

(1+cos(12n)2)n(1+12)n=1
Now cosu1u22 for small u0, then
(1+cos(12n)2)n(1122(n+1))n1
movingsupplyw1

movingsupplyw1

Beginner2022-01-04Added 30 answers

Continuing where you left off with LHopitals:
L=limnnln(cos(12n+1))=limnln(cos(2n1))1n=limntan(2n1)2n1ln21n2
Supposing the limits are finite, we can break up the limit over multiplication like this:
ln2limntan(2n1)limnn22n+1
The first limit is 0 by the continuity of tan, and the second is easily shown to be 0 by repeatedly applying LHopitals.
So, our original limit is e2L=e0=1
Vasquez

Vasquez

Expert2022-01-08Added 669 answers

You were very close to the right track. Write
y=(1+cos(ϵn)2)nlog(y)=nlog(1+cos(ϵn)2)
Now, using the series expansion of cos(t) for small angles (or equivalents
cos(ϵn)=112ϵ2n+
1+cos(ϵn)2=114ϵ2n+
log(1+cos(ϵn)2)=14ϵ2n+
But ϵ=12 ; so
log(y)0y1

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