The infinity norm of the sequence $v(n) = n \sin(n!)/(n^2+1)$

Question

For a bounded sequence $v(n)$, $n\in\mathbb{Z}$ define $$||v||_\infty = \max_{n\in\mathbb{Z}} |v(n)|.$$ Let $$v(n) =\frac{n\sin{(n!)}}{n^2+1},$$ and find whether $||v||_\infty<\infty$.

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\begin{align} ||v||_\infty &= \max_{n\in\mathbb{Z}}\left|\frac{n\sin{(n!)}}{n^2+1}\right| \\ &\le \max_{n\in\mathbb{Z}}\left|\frac{n}{n^2+1}\right|\cdot \max_{n\in\mathbb{Z}}\left|sin{(n!)}\right| \\ &\le \frac{1}{2}\cdot1 = \frac{1}{2} <\infty. \end{align}

So indeed $||v||_\infty$ is finite.

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Does this seem like a legal method to solving this question?

Answer 1

As Farnight suggested, it is better to use $\sup$. I also prefer to work on a more concrete level: with numbers, not with sets of numbers. So I would write: "for $n\in \mathbb{N} $, $$|v(n)|=\left|\frac{n\sin{(n!)}}{n^2+1}\right|\le \frac{n }{n^2+1}\le \frac{n^2}{n^2+1} <1 $$ Therefore, the sequence $v$ is bounded."

Note that your write up leaves the part $$\max_{n\in\mathbb{Z}}\left|\frac{n}{n^2+1}\right| \le\frac12$$ unexplained, while it is arguably the most substantial part of the computation.