Frederick Kramer

Answered

2022-07-07

Find all values of $\alpha $ such that series

$\sum _{n=1}^{\mathrm{\infty}}{(\frac{1}{n\cdot \mathrm{sin}(1/n)}-\mathrm{cos}\left(\frac{1}{n}\right))}^{\alpha}$

converges.

I used Maclaurin for sin and cos and got:

${a}_{n}={(\frac{1}{1-{\displaystyle \frac{1}{3!{n}^{2}}}+\dots}-1+\frac{1}{2!{n}^{2}}-\frac{1}{4!{n}^{4}}+\dots )}^{\alpha}$

Put it together in one fraction seems to be a hard thing to do.

$\sum _{n=1}^{\mathrm{\infty}}{(\frac{1}{n\cdot \mathrm{sin}(1/n)}-\mathrm{cos}\left(\frac{1}{n}\right))}^{\alpha}$

converges.

I used Maclaurin for sin and cos and got:

${a}_{n}={(\frac{1}{1-{\displaystyle \frac{1}{3!{n}^{2}}}+\dots}-1+\frac{1}{2!{n}^{2}}-\frac{1}{4!{n}^{4}}+\dots )}^{\alpha}$

Put it together in one fraction seems to be a hard thing to do.

Answer & Explanation

Franco Cohen

Expert

2022-07-08Added 8 answers

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

$\frac{1}{n\mathrm{sin}(1/n)}-\mathrm{cos}(1/n)=\frac{2}{3{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)$

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

$\frac{1}{n\mathrm{sin}(1/n)}-\mathrm{cos}(1/n)=\frac{2}{3{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)$

skynugurq7

Expert

2022-07-09Added 3 answers

Hint

$n\mathrm{sin}(\frac{1}{n})=1-\frac{1}{6{n}^{2}}(1+{\u03f5}_{1}(n))$

$\frac{1}{n\mathrm{sin}(\frac{1}{n})}=1+\frac{1}{6{n}^{2}}(1+{\u03f5}_{2}(n))$

$\mathrm{cos}(\frac{1}{n})=1-\frac{1}{2{n}^{2}}(1+{\u03f5}_{3}(n))$

thus, when $n\to +\mathrm{\infty}$, the general term of your series ${u}_{n}$, satisfies

${u}_{n}\sim (\frac{2}{3{n}^{2}}{)}^{\alpha}$

and by the limit comparison test,

$\sum {u}_{n}$ converges $\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\alpha >\frac{1}{2}$

$n\mathrm{sin}(\frac{1}{n})=1-\frac{1}{6{n}^{2}}(1+{\u03f5}_{1}(n))$

$\frac{1}{n\mathrm{sin}(\frac{1}{n})}=1+\frac{1}{6{n}^{2}}(1+{\u03f5}_{2}(n))$

$\mathrm{cos}(\frac{1}{n})=1-\frac{1}{2{n}^{2}}(1+{\u03f5}_{3}(n))$

thus, when $n\to +\mathrm{\infty}$, the general term of your series ${u}_{n}$, satisfies

${u}_{n}\sim (\frac{2}{3{n}^{2}}{)}^{\alpha}$

and by the limit comparison test,

$\sum {u}_{n}$ converges $\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\alpha >\frac{1}{2}$

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