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I am having problems showing either convergence or divergence of the series mentioned in the heading. I've tried all techniques which I know of, divergence-test, ratio-test, limit comparison-test but I have been unable to solve it. I have however tried using power series to solve it and I will show my attempt here. My question otherwise is how do I solve this and is my method of using power series correct?

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

$$\sum_{k=1}^{\infty}\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left(\frac{1}{k}\right)^{2n+1}-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\left(\frac{1}{k}\right)^{2n+1}\right) =$$

$$\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{k}\right)^{2n+1}\left(\frac{1}{(2n+1)!}-\frac{1}{(2n+1)} \right)=$$

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{(2n+1)!}-\frac{1}{(2n+1)} \right)+\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{2}\right)^{2n+1}\left(\frac{1}{(2n+1)!}-\frac{1}{(2n+1)} \right)+\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{3}\right)^{2n+1}\left(\frac{1}{(2n+1)!}-\frac{1}{(2n+1)} \right)+ \dots$$

For an arbitrary value of $k=\alpha$ a ratio test will yield that the radius of convergence is $R = \infty$ therefore all of the sums converge and thus the original series converges. I am not certain that this is a sufficient argument to conclude that the series is converging, but if you have another method I am glad to see that.

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$\sin(1/k) = 1/k + 1/6k^3 + o(1/k^4)$

$\arctan(1/k) = 1/k - 1/3k^3 + o(1/k^4)$

so $\sin(1/k) - \arctan(1/k) = 1/2k^3 + o(1/k^4)$

By Riemann sum comparison, the series converges.

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  • $\begingroup$ How do we know rigorously that the sum of $O(\frac{1}{k^4})$ converges? $\endgroup$ – Sindbad Mar 1 '18 at 16:00
  • $\begingroup$ @Sindbad: Let $u_k$ be this $o(1/k^4)$. For k large enough, $0 ≤ |u_k| ≤ 1/k^4$. $\sum 1/k^4$ converges hence $\sum u_k$ converges absolutely hence converges. $\endgroup$ – user371663 Mar 1 '18 at 16:06
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Let $f(x)=\sin x-\arctan x$. Then by Lagrange's MVT, one has $$ f(\frac{1}{k})-f(0)=\frac{1}{k}f'(\xi_k), \xi_k\in(0,\frac1k).$$ Noting \begin{eqnarray} |f'(x)|&=&\bigg|\cos x-\frac1{1+x^2}\bigg|\\ &\le&|1-\cos x|+\bigg|1-\frac1{1+x^2}\bigg|\\ &=&2\sin^2(\frac{x}{2})+\frac{x^2}{1+x^2}\\ &\le&\frac{1}{2}x^2+x^2\\ &=&\frac32x^2, \end{eqnarray} one has $$ |f(\frac1k)|=\frac1k|f'(\xi_k)|\le\frac32\frac{1}{k^3}$$ and hence the series converges.

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Better idea: use Taylor to approximate $$\sin(\frac{1}{k})-\arctan(\frac{1}{k}) = \cdots.$$

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This sort of thing is easy once you master big-O notation. For all $x$, $$\sin x=x-\frac{x^3}6+\frac{x^5}{120}-\cdots$$ and if $|x|<1$, $$\arctan x=x+\frac{x^3}3+\frac{x^5}6+\cdots.$$ Thus $\sin x=x+O(x^3)$ and $\arctan x=x+O(x^3)$ as $x\to0$, that is $|\sin x-x|\le C|x|^3$ etc., on an interval around zero. Therefore $\sin x-\arctan x=O(x^3)$ and so $\sin (1/k)-\arctan (1/k)=O(1/k^3)$. As $\sum_k 1/k^3$ converges then so does $\sum_k(\sin (1/k)-\arctan (1/k))$, absolutely.

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Both $\sin x$ and $\arctan x$ are odd, analytic functions in a neighbourhood of the origin, behaving like $x$ as $x\to 0$. It follows that $\sin(x)-\arctan(x)=O(x^3)$ as $x\to 0$ and the given series is convergent by asymptotic comparison with $\sum_{k\geq 1}\frac{1}{k^3}=\zeta(3)$.

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