Frederick Kramer

2022-07-07

Find all values of $\alpha$ such that series
$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n\cdot \mathrm{sin}\left(1/n\right)}-\mathrm{cos}\left(\frac{1}{n}\right)\right)}^{\alpha }$
converges.
I used Maclaurin for sin and cos and got:
${a}_{n}={\left(\frac{1}{1-\frac{1}{3!{n}^{2}}+\dots }-1+\frac{1}{2!{n}^{2}}-\frac{1}{4!{n}^{4}}+\dots \right)}^{\alpha }$
Put it together in one fraction seems to be a hard thing to do.

Franco Cohen

Expert

HINT:
You are on the right track. Note that
$\frac{1}{1-\frac{1}{6{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)}=1+\frac{1}{6{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)$
Hence, we see that
$\frac{1}{n\mathrm{sin}\left(1/n\right)}-\mathrm{cos}\left(1/n\right)=\frac{2}{3{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)$

skynugurq7

Expert

Hint
$n\mathrm{sin}\left(\frac{1}{n}\right)=1-\frac{1}{6{n}^{2}}\left(1+{ϵ}_{1}\left(n\right)\right)$
$\frac{1}{n\mathrm{sin}\left(\frac{1}{n}\right)}=1+\frac{1}{6{n}^{2}}\left(1+{ϵ}_{2}\left(n\right)\right)$
$\mathrm{cos}\left(\frac{1}{n}\right)=1-\frac{1}{2{n}^{2}}\left(1+{ϵ}_{3}\left(n\right)\right)$
thus, when $n\to +\mathrm{\infty }$, the general term of your series ${u}_{n}$, satisfies
${u}_{n}\sim \left(\frac{2}{3{n}^{2}}{\right)}^{\alpha }$
and by the limit comparison test,
$\sum {u}_{n}$ converges $\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\alpha >\frac{1}{2}$

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