# Find the radius of convergence for infinite series. [duplicate]

I have just shown via the Taylor expansion for $$\sin(\frac{1}{n})$$ that the series $$\sum_{n=1}^{\infty}\left(\frac{1}{n} - \sin\left(\frac{1}{n}\right)\right)$$

is in fact convergent and now I'm asked to find the radius of convergence for the series

$$\sum_{n=1}^{\infty}\left(\frac{1}{n} - \sin\left(\frac{1}{n}\right)\right)\cdot x^n ,\; x\in\mathbb{R}$$

and I'm not sure how to approach this. I have tried the ratio test but it doesn't go well and the Cauchy-Hadamard doesn't seem to work either.

Any ideas?

As $$\frac1n-\sin\frac1n\sim \frac1{6n^3}$$, we have $$\frac1R=\limsup_{n\to\infty}\sqrt[n]{\frac1n-\sin\frac1n}=\limsup_{n\to\infty}\sqrt[n]{\frac1{6n^3}}=1$$
Use Limit Comparison: $$1/n - \sin(1/n)= O(1/n^3)$$. The corresponding series $$\sum_{n=1}^{\infty}\frac{1}{n^3}x^n$$ has radius of convergence $$1$$ and converges everywhere on the boundary; your series has the same behavior.