# Radius of convergence and uniformly convergence

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?

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$$(\forall n\in\Bbb N):\left|\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}\right|Then you have, if $$|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}$$ converges.
• Done. ${}{}{}{}$ – José Carlos Santos May 29 at 11:37
• is the proof to the first limit $\lim_{n\to\infty}\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}=\frac16$ very trivial ? Am i missing something very trivial? – Noob mathematician May 30 at 21:08
• Since$$\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.$$ – José Carlos Santos May 30 at 21:11