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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?

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As $\frac1n-\sin\frac1n\sim \frac1{6n^3}$, we have $$\frac1R=\limsup_{n\to\infty}\sqrt[n]{\frac1n-\sin\frac1n}=\limsup_{n\to\infty}\sqrt[n]{\frac1{6n^3}}=1 $$

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  • $\begingroup$ Ok so the Cauchy-Hadamard did in fact work, thanks. $\endgroup$ – Eod Enaj May 30 at 21:25
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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.

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