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I have to find the value of a such that the following series converges:$$\sum_{n=1}^{\infty}n^{\frac{1}{3}}\left|\sin\left(\frac{1}{n}\right)-\frac{1}{n^a}\right|$$

First of all, I simplified the series with the respective asymptotic as follows: $$\sum_{n=1}^{\infty}n^{\frac{1}{3}}\left|\frac{1}{n}-\frac{1}{n^a}\right| \sim \sum_{n=1}^{\infty}n^{\frac{1}{3}}\left|\frac{1}{n^a}\right|$$ Then, simplifying again, it results: $$\sum_{n=1}^{\infty}\frac{1}{n^{a-\frac{1}{3}}}$$ Which should converge for $a>\frac{4}{3}$, but I checked with Mathematica, which according to it, it is wrong.

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    $\begingroup$ Uhm... Are you sure $\left\lvert \frac1n-\frac1{n^\alpha}\right\rvert\sim \frac1{n^\alpha}$? That ain't looking right. $\endgroup$ – Saucy O'Path Apr 13 at 15:38
  • $\begingroup$ As @SaucyO'Path mentions, $\frac{1}{n} > \frac{1}{n^a}$ for all $a > 1$. Therefore, you could make a distinction into two cases, $a > 1$ and $a > 1$ and then you could get rid of the absolute value. $\endgroup$ – Viktor Glombik Apr 13 at 15:39
  • $\begingroup$ Mathematica is like a gun. Sure, it comes in handy every now and then, but it ain't no smarter than the dude using it. $\endgroup$ – Ivan Neretin Apr 13 at 15:40
  • $\begingroup$ Also, a very special thing happens at $a=1$. $\endgroup$ – Ivan Neretin Apr 13 at 15:42
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The first step is wrong.

(First of all, $a_n\sim b_n$ and $\sum_{n=1}^\infty a_n \sim \sum_{n=1}^\infty b_n$ are not the same thing; the second, actually, does not mean anything. But that's not the main point...)

Note that, as $n\to\infty$, $$ \sin\frac{1}{n} - \frac{1}{n^a} = \frac{1}{n} - \frac{1}{6n^3} - \frac{1}{n^a} + o\left(\frac{1}{n^3}\right) $$ so that $$ n^{1/3}\left(\sin\frac{1}{n} - \frac{1}{n^a} \right) = \left(\frac{1}{n^{2/3}} - \frac{1}{n^{a-1/3}} \right) - \frac{1}{6n^{8/3}} + o\left(\frac{1}{n^{8/3}}\right) $$ So for the series $\sum_n n^{1/3}\left|\sin\frac{1}{n} - \frac{1}{n^a} \right|$ to converge, you need the term in parentheses to cancel (why?), which gives you what $a$ must be.

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  • $\begingroup$ Perfect answer, thank you very much! $\endgroup$ – Kevin Apr 13 at 16:22
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    $\begingroup$ You're welcome! $\endgroup$ – Clement C. Apr 13 at 16:22
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If you take $a=1$, then $\left\lvert\sin\left(\frac1n\right)-\frac1n\right\rvert$ behaves as $\frac1{n^3}$, since$$\lim_{n\to\infty}\frac{\left\lvert\sin\left(\frac1n\right)-\frac1n\right\rvert}{\frac1{n^3}}=\lim_{x\to0^+}\frac{\left\lvert\sin(x)-x\right\rvert}{x^3}=\frac16.$$So, take $a=1$ and use the fact that the series$$\sum_{n=1}^\infty\frac{\sqrt[3]n}{n^3}$$converges.

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