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convergence of $ \sum_{k=1}^\infty \sin^2(\frac 1 k)$

How can I do this? Should I use the Ratio Test (I tried this but it started getting complicated so I stopped)? Or the Comparison test(what should I compare it to?)?

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    $\begingroup$ Comparison or Limit comparison with $\sum \frac{1}{k^2}$. The Ratio Test won't work, the convergence of the terms to $0$ is too slow for that. $\endgroup$ Jul 12 '15 at 1:25
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Comparison: $ \displaystyle \left(\sin \frac 1 k \right)^2 < \left( \frac 1 k \right)^2$.

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  • $\begingroup$ How did you derive that inequality? $\endgroup$
    – TanMath
    Jul 12 '15 at 1:27
  • $\begingroup$ For $x\in (0,\pi/2)$, define $g(x)=x-\sin x$. Then, we have $g'(x)=1-\cos x >0$ for every $x$ in that interval. That means that the function is strictly increasing, and $g(0)=0$, so we have $x-\sin x>0 \implies \sin x<x$. Note that $1/k < \pi/2$ for every $k\geq 1$. $\endgroup$ Jul 12 '15 at 1:35
  • $\begingroup$ @TanMath : Notice that $\sin'(0)=1$ but $\sin'x = \cos x<1$ for $0<x\le1$. The mean value theorem says $\Big( \sin x - \sin 0\Big)/\Big(x-0\Big) = \sin' c = \cos c < 1$ for some $c$ between $0$ and $x$. Hence $\sin x \cdot \cos c = x$, so $\sin x < x$. (This assumes $x>0$, since multiplying both sides by $x$ would otherwise require changing "$<$" to "$>$". That is a rigorous solution, but you should also know another approach: Think about what the graph of sine function looks like, with its slope of $1$ at $x=0$, and what the line $y=x$ looks like. That should tell you${}\,\ldots\qquad{}$ $\endgroup$ Jul 12 '15 at 1:35
  • $\begingroup$ $\ldots\,{}$that the curve lies below the line, so $\sin x<x$ whenever $x>0$. Your understanding of the sine-in-radians function is incomplete if you haven't got that. ${}\qquad{}$ $\endgroup$ Jul 12 '15 at 1:36
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    $\begingroup$ So $x=1/k$. When I say $\sin x<x$ whenever $x>0$, that's saying something about ALL positive numbers. And $1/k$ is a positive number. Whatever applies to ALL positive numbers applies to $1/k$. ${}\qquad{}$ $\endgroup$ Jul 12 '15 at 1:40
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With equivalents: $\sin^2\dfrac1k\sim_\infty \dfrac1{k^2}$. The series $\displaystyle\sum\dfrac1{k^2}$ converges absolutely, hence $\displaystyle\sum\sin^2\dfrac1k$ converges absolutely.

I recall $f(x)\sim_\infty g(x)$, for two functions defined in a neighbourhood of $\infty$ roughly means $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$$ To take into account that $g(x)$ can be equal to $0$, there is a technical definition: $$f(x)-g(x)=_{\infty}o\bigl(g(x)\bigr),$$ which is equivalent to the previous definition in case $g(x)\neq 0$ in some neighbourhood of $\infty$.

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  • $\begingroup$ I think you ought to say what $\sim_\infty$ means when posting something like this. Is this the same as limit comparison? ${}\qquad{}$ $\endgroup$ Jul 12 '15 at 1:52
  • $\begingroup$ Don't you think it's clear from emphasising the word ‘equivalent’? $\endgroup$
    – Bernard
    Jul 12 '15 at 2:03
  • $\begingroup$ No. There are many equivalence relations. ${}\qquad{}$ $\endgroup$ Jul 12 '15 at 2:05
  • $\begingroup$ OK, I'll dot it. Buts as far as I know there is only one sense for the noun ‘equivalent’ $\endgroup$
    – Bernard
    Jul 12 '15 at 2:07
  • $\begingroup$ There are certainly many senses of the word "equivalent". ${}\qquad{}$ $\endgroup$ Jul 12 '15 at 12:11

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