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The assignment that I am having trouble with is as follows:

a) Use the fact that $\lim_{n\rightarrow\infty}n^3\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)=\frac{1}{6}$ to show that

$$\sum_{n=0}^\infty\left(\frac 1n-\sin\left(\frac 1n\right)\right)$$

converges.

b) Determine the radius of convergence of the power series $$\sum_{n=0}^\infty\left(\frac 1n-\sin\left(\frac 1n\right)\right)x^n$$

I am having real trouble with computing the limit $$\lim_{n\rightarrow\infty}\left\lvert\frac 1n-\sin\left(\frac 1n\right)\right\rvert^{\frac1n}=r^{-1}$$ Here r is the radius of convergence.

An0ku
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2 Answers2

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As the first part (to use limit comparison test then) was given in this answer (indeed $\lim\limits_{n\to\infty}\frac{\frac{1}{n}-\sin\left(\frac{1}{n}\right)}{\frac{1}{n^3}}$ is given to be $\frac{1}{6}$), the second: $\frac{1}{n}=x$
(that's an only a variable change for the limit, no relation to the $x$ in the question), $$\begin{align*}\lim\limits_{x\to 0}|x-\sin x|^x&= \lim\limits_{x\to 0}\left|\frac{x-\sin x}{x^3}\right|^x\cdot x^{3x}\\ &=\lim\limits_{x\to 0}\left|\frac{1}{6}\right|^x\cdot e^{3x\ln x}=1 \end{align*}$$ Where $\lim\limits_{x\to 0} x \ln x$ is known to be $0$.

Alexey Burdin
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  • Thank you for linking that page. I am assuming that by the comparison test 1/n - sin(1/n) < 1/(6n^3) therefore 1/n - sin(1/n) converges. Can you explain why that inequality holds? – An0ku Jun 03 '20 at 18:01
  • Please see the edits. Questions welcome, if any. – Alexey Burdin Jun 03 '20 at 18:24
  • Thank you very much for taking the time to help. It is much appreciated – An0ku Jun 03 '20 at 18:45
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$$\sin(\frac 1n)=\frac 1n -\frac{1}{6n^3}(1+\epsilon(n))$$ with $$\lim_{n\to+\infty}\epsilon(n)=0$$ thus

$$u_n= \frac 1n-\sin(\frac 1n)\sim \frac{1}{6n^3}$$

the two series $\sum u_n x^n$ and $ \sum \frac{1}{n^3} x^n$ will have the same radius of convergence $(R=1)$.

hamam_Abdallah
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