Question

Investigate the convergence of the series
`\lim_(n \to \infty) sin(n)`

Answer 1

The series
`\lim_(n \to \infty) sin(n)` does not converge, because the function
`sin(n)` oscillates infinitely between -1 and 1 as n approaches infinity.

More formally, for any positive integer
`k`, we can find an integer
`n` such that
`n\pi` is very close to
`\pi`, which implies that
`sin(n)\approx(-1)^k`. Therefore, the sequence
`{sin(n)}` does not converge to any limit as
`n` approaches infinity.

Alternatively, we can also prove that
`\lim_(n \to \infty) sin(n)` does not exist using the epsilon-delta definition of a limit. Suppose there exists a limit
`L`, then for any
`\epsilon > 0`, there exists a positive integer
`N` such that for all
`n > N`, we have
`|sin(n) - L| < \epsilon`.

However, we can choose
`\epsilon = 1/2` and find two subsequences
`{a_n}` and
`{b_n}` such that
`|sin(a_n) - sin(b_n)| = 1`. Therefore, for any possible limit
`L`, we can always find two subsequences that converge to different values, contradicting the assumption that the limit exists.

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