Calculate \lim_{n\to\infty}|\sin(\pi\sqrt{n^2+n+1})|

Kelvin Gregory

Kelvin Gregory

Answered question

2022-02-27

Calculate
limn|sin(πn2+n+1)|

Answer & Explanation

Rebekah Irvine

Rebekah Irvine

Beginner2022-02-28Added 6 answers

Note that
n2+n+1n=n+1n2+n+1+n12
as n
For even n, sin(n2+n+1π)=sin(n2+n+1πnπ)sin(π2) as n, n even.
For odd n, sin(n2+n+1π)=sin(n2+n+1πnπ)sin as n, n odd.
Therefore,
|sin(n2+n+1π)|1
as n.

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