Exploring the Convergence of the Sequence (a_n n sin(n))

When analyzing whether a sequence like (a_n n sin(n)) has a convergent subsequence, we need to delve into the detailed behavior of the components involved.

Boundedness of (sin(n))

The function (sin(n)) oscillates between -1 and 1 for all real numbers (n). Therefore, for the sequence (a_n n sin(n)), we have the inequality:

[ -n leq n sin(n) leq n ]

Behavior of (n sin(n))

As (n) increases, the term (n) grows without bound while (sin(n)) continues to oscillate between -1 and 1. Consequently, (n sin(n)) will also grow without bound in a fluctuating manner, due to the oscillatory nature of the sine function.

Can We Find a Convergent Subsequence?

Despite the unbounded growth of (n sin(n)), we can exploit the oscillatory nature of (sin(n)) to find a convergent subsequence. Notably, (sin(n)) takes values densely in the interval ([-1, 1]) as (n) varies over the integers. This suggests that there exist indices (n_k) such that (sin(n_k)) approaches some value as (k) increases.

Example of a Convergent Subsequence

For instance, to find a convergent subsequence that converges to 0, we can choose (n_k kpi) where (k) is an integer. When (n_k kpi), (sin(n_k) 0), and thus the subsequence (a_{n_k}) is:

[ a_{n_k} n_k sin(n_k) kpi cdot 0 0 ]

This subsequence converges to 0, indicating that the sequence (a_n n sin(n)) indeed has a convergent subsequence.

Alon Amit's Analysis

Alon Amit provided an interesting insight into why the sequence (n sin(n)) may not have a subsequence converging to 0. He explored the existence of a subsequence (n_{k_0}) such that:

[ lim_{k to infty} frac{n_{k_0} sin(n_{k_0})}{n_{k_0}} epsilon_0 ]

This analysis led to the inequality:

[ -frac{epsilon_0}{n_k} cos(m_kpi) leq frac{n_k sin(n_k)}{n_k} leq frac{epsilon_0}{n_k} ]

Expanding (sin(n_k)) using a Taylor series near (m_kpi), we get:

[ sin(f_k) approx f_k - frac{f_k^3}{6} O(f_k^5) ]

where (f_k n_k - m_kpi) and (|f_k| leq frac{pi}{2}). Simplifying, we find:

[ frac{epsilon_0}{n_k} approx f_k approx frac{epsilon_0}{n_k^3} ]

This implies that:

[ frac{n_k}{m_k} - pi frac{epsilon_0}{m_k^2} approx frac{epsilon_0}{n_k^3} ]

If such a subsequence existed, it would imply that the irrationality measure of (pi) is strictly greater than 2. However, as of 2023, the best known inequality for the irrationality measure (mu) of (pi) is (2 leq mu leq 7.10320533413).

The best known inequality is:

[ frac{n_k}{m_k} - pi frac{1}{m_k^2 sqrt{5}} ]

which implies that:

[ n_k sin left(frac{n_k}{sqrt{5}}right) approx 1.40496 ]

for infinitely many positive integers (n_k). This does not necessarily mean that the interval ([- frac{pi}{sqrt{5}}, frac{pi}{sqrt{5}}]) is visited infinitely many times by numbers of the form (n_k sin(n_k)). In fact, the distribution of (n sin(n)) is highly non-trivial, and finding integers (n) such that (|n sin(n)|

Conclusion

The sequence (a_n n sin(n)) does have a convergent subsequence, as demonstrated by the subsequence (n_k kpi) converging to 0. However, further analysis suggests that the irrationality measure of (pi) and the distribution of (n sin(n)) make it challenging to find a subsequence converging to a value significantly smaller than the irrationality measure of (pi).