For what values of $p$ does the sequence $a_n=\frac{sin(1/n)}{n^{10}}$ belong to $l^p$ space?

Question

For what values of $p$ does the sequence $a_n=\frac{sin(1/n)}{n^{10}}$ belong to $l^p$ space?

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If $p=1$ then we need to check the convergence of the series $\sum_{n=1}^{\infty}|a_n|$. To use the ratio test, we have

$\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=...=\lim_{n\to\infty}\frac{1}{n}|\frac{\sin(1/(n+1))}{\sin(1/n)}|$ But how can I determine whether the limit is less than $1$ or not?

In fact for the general case, we need to check whether the sequence $b_k=\sum_{n=1}^{k}|\frac{sin(1/n)}{n^{10}}|^p$ is convergent where $0<p$. Thanks.

Answer 1

To simplify things, note that we have

$$\lim_{x\to 0} \frac{\sin x}{x} = 1,$$

thus $\sin \frac 1n $ is more or less $\frac{1}{n}$ when $n$ is large and so

$$\frac{\sin \frac 1n}{n^{10} }\sim \frac{1}{n^{11}}.$$

Now to check if $a_n$ is in $\ell^p$, it suffices to know when will

$$\sum \left( \frac{1}{n^{11}}\right)^p = \sum \frac{1}{n^{11p}}$$

converges. Note that this converges precisely when $11p >1$.

Answer 2

$L^p$ spaces are defined only for $p\in [1,+\infty)$. Anyway the only thing to notice is that $|sin(\frac{1}{n})|$ is asyntotic to $|\frac{1}{n}|$ for $n\rightarrow +\infty$. You can then say that $\sum_0^ \infty |\frac{sin(\frac{1}{n})}{n^{10}}|^p$ convereges if $\sum_0^ \infty |\frac{1}{n^{11}}|^p$ does. This happens for all $p \in [1,+\infty)$.