I'm trying to prove whether or not $$ \sum_{x = 1}^\infty x^2 \left (\cos\left (\frac{1}{x}\right ) -1 \right ) $$ converges. Based on graphs, I think that the sequence $\{x^2 (\cos (1/x) - 1)\}$ does not converge to $0,$ so once I prove that, I know that the series above diverges. But, I'm not sure how to show that the sequence converges to $0.$
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3 Answers
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You are right, it does not converge to $0$. Let $t=\frac{1}{x}$. We show that $$\frac{\cos t-1}{t^2}\tag{1}$$ does not converge to $0$ as $t$ approaches $0$ from the right. Multiply top and bottom by $-(1+\cos t)$. We get that for $t\ne 0$, (1) is equal to $$-\frac{1}{1+\cos t}\cdot\frac{\sin^2 t}{t^2}.$$ The rest is undoubtedly familiar. As $t\to 0^+$, the function (1) converges to $-\frac{1}{2}$.
answered Oct 1, 2014 at 6:44
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Hint: Maclaurin series of $\cos$.
answered Oct 1, 2014 at 6:44
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Using the Maclaurin expansion $$\cos\left (\epsilon\right )=1-\frac{\epsilon^2}{2}+O(\epsilon^2)$$ you get, as $x$ tends to infinity,$$x^2 \left (\cos\left (\frac{1}{x}\right ) -1 \right )=x^2\cdot\left(-\frac{1}{2\:x^2}\right)+O\left(\frac{1}{x^2}\right ) =-\frac{1}{2}+O\left(\frac{1}{x^2}\right )$$ and the considered series does not converge.
