This is a question we asked on a second semester calculus test.
For what values of $p$ does this series converge? $$\sum_{n=1}^{\infty}\frac{\sin(1/n)}{n^p}$$
I believe that it actually can be shown that $p> 0$ is a valid answer.
However. I am interested in finding a proof that is simple enough that a beginning calculus student could do on their own cognition. Is there a simple way to give the exact values of $p$ for which the series converges?
$\endgroup$ 24 Answers
$\begingroup$Limit comparison test.
$$\lim \frac{\sin(1/n)/n^p}{1/n^{p+1}}=1,$$
$\sum\frac{1}{n^{p+1}}$ converges when $p>0$, diverges when $p\leq 0$.
$\endgroup$ $\begingroup$Since $$ \sin \frac1n \approx \frac1n, $$ your series is equivalent to $$ \sum_n \frac{1}{n^{p+1}}. $$
$\endgroup$ $\begingroup$$$ \frac{\sin 1/n}{n^p} \sim \frac{1/n}{n^{p}} \,\,\, \text{if }n\to \infty $$ so I guess every $p>0$ :)
$\endgroup$ $\begingroup$Comparison test:
- Note that $$-1 \leq \sin(x) \leq 1 ~~ \forall ~x$$
- Therefore, $$\dfrac{\sin(x)}{f(n)} \leq \dfrac{1}{f(n)} ~~ \forall ~x$$
So for your case: $f(n)=n^p$. This converges by $p$-series when $p>1$. Therefore, your original series converges for the same.
You cannot use alternating series test, since: $$\lim_{n\rightarrow \infty} \frac{1}{n} \rightarrow 0^+$$ Thus, the series ends up being strictly positive for large enough $n$.
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