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I'm trying to prove this series converges by using some sort of comparison test.

$$\sum_{n=1}^{\infty}\frac{1}{n^{0.51}}-\sin\left(\frac{1}{n^{0.51}}\right)$$

I know by $\sin n\le n$ that the series is positive, so I went with the direction of using the comparison test. But I can't seem to find a function that is always greater than the expression in the series that also converges..

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3 Answers 3

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Using the Taylor series for $\sin x$, we have $$\sin \frac{1}{n^p} = \frac{1}{n^p} + O\left(\frac{1}{n^{2p}}\right)$$ Therefore $$S=\sum_{n\ge 1}\frac{1}{n^p} - \sin \frac{1}{n^p} = \sum_{n\ge 1}\frac{1}{n^p} - \frac{1}{n^p} + O\left(\frac{1}{n^{2p}}\right) = \sum_{n\ge 1}O\left(\frac{1}{n^{2p}}\right)$$ Hence, the series $S$ converges if $p > 0.5$.

Your series converges since $0.51 > 0.5$.

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    $\begingroup$ Hi thanks for the response! We haven't gone through Taylor Series yet unfortunately. :( $\endgroup$ May 10, 2021 at 13:11
  • $\begingroup$ The series converges iff $p\gt\frac13$. What you've shown is that the series converges if $p\gt\frac12$. $\endgroup$
    – robjohn
    May 10, 2021 at 22:20
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You can use the fact that for $x \geq 0$, $$x - \sin(x) \leq \frac{1}{6} x^3.$$

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  • $\begingroup$ How would one prove such a claim? $\endgroup$ May 10, 2021 at 13:12
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    $\begingroup$ @MathCurious Show that $f(x) = \frac16 x^3 - x + \sin x \ge 0$ $\endgroup$
    – VIVID
    May 10, 2021 at 13:21
  • $\begingroup$ So how do you prove that the RHS function converges? $\endgroup$ May 10, 2021 at 14:05
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    $\begingroup$ Yes, that is what I'm asking $\endgroup$ May 10, 2021 at 14:28
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    $\begingroup$ @MathCurious: this answer gives a pre-calculus proof that $\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$. $\endgroup$
    – robjohn
    May 10, 2021 at 21:04
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Using Heine's definition for limits at $\infty$ and using the limit comparison test with $\;b_n=\frac{1}{6\left(n^{0.51}\right)^{3}}=\frac{1}{6n^{1.53}}$ , we get:

$$\lim_{n\to\infty}\frac{\frac{1}{n^{0.51}}-\sin\left(\frac{1}{n^{0.51}}\right)}{\frac{1}{6n^{1.53}}}{=}\lim_{x\to\infty}\frac{\frac{1}{x^{0.51}}-\sin\left(\frac{1}{x^{0.51}}\right)}{\frac{1}{6x^{1.53}}}\underset{t=\frac{1}{x^{0.51}}}{=}\lim_{t\to0^{+}}\frac{t-\sin t}{\frac{t^{3}}{6}}\underset{L}{=}$$ $$\lim_{t\to0^{+}}\frac{1-\cos t}{\frac{t^{2}}{2}}\underset{L}{=}\lim_{t\to0^{+}}\frac{\sin t}{t}=1$$ Hence $\displaystyle{\sum_{n=1}^{\infty}\left[\frac{1}{n^{0.51}}-\sin\left(\frac{1}{n^{0.51}}\right)\right]}$ converges with $\displaystyle{{\displaystyle \sum_{n=1}^{\infty}b_{n}}}$.

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