Radius of convergence and uniformly convergence

Question

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?

Answer 1

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.