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I think the radius of convergence for $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$, $x\in \mathbb R$ is:

$r^{-1}$=$\lim_{n\to \infty}$$|\frac{a_{n+1}}{a_n}$|=1 so we get that $r$=1.

But how can I show it formally?

After that, I have to show that $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$ is uniformly convergent in the interval $[-r,r]$. Can I maybe use Weierstrass majoranttest? What can I use as majorant serie?

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Since$$\lim_{n\to\infty}\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}=\frac16$$and since the radius of convergence of the power series $\sum_{n=1}^\infty\frac{z^n}{n^3}$ is $1$, it follows from the comparaison test that the radius of convergence of your series is $1$ too.

By the same argument, together with the help of the Weierstrass $M$ test, it follows that the series converges uniformly on $\overline{D(0,1)}$. In fact, since the sequence$$\left(\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}\right)_{n\in\Bbb N}$$converges, it is bounded. Take $M\in(0,\infty)$ such that$$(\forall n\in\Bbb N):\left|\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}\right|<M.$$Then you have, if $|z|\leqslant1$,$$\left|\left(\frac1n-\sin\left(\frac1n\right)\right)z^n\right|\leqslant\frac M{n^3}|z^n|\leqslant\frac M{n^3}$$and the series $\sum_{n=1}^\infty\frac M{n^3}$ converges.

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  • $\begingroup$ Thank you it makes a bit sence. Can you maybe write a bit more about the argument for uniformly convergent by Weierstrass 𝑀 test? $\endgroup$ – Lifeni May 29 at 11:28
  • $\begingroup$ Done. ${}{}{}{}$ $\endgroup$ – José Carlos Santos May 29 at 11:37
  • $\begingroup$ Thank you. Now it makes sense $\endgroup$ – Lifeni May 29 at 12:00
  • $\begingroup$ is the proof to the first limit $\lim_{n\to\infty}\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}=\frac16$ very trivial ? Am i missing something very trivial? $\endgroup$ – Noob mathematician May 30 at 21:08
  • $\begingroup$ Since$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,$$you have$$\lim_{x\to0}\frac{\sin(x)-x}{x^3}=\lim_{x\to0}\frac{-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots}{x^3}=-\frac16.$$ $\endgroup$ – José Carlos Santos May 30 at 21:11

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