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I am wondering how to determine whether the following series converges using the comparison test:

$$\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$$

In class, my professor used the idea that $\frac{\sin x}{x}\to 1$ as $x$ gets large.

So, my professor said that since $$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}\to\frac{2}{3}$$by the comparison test, $\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$ converges. Can someone explain why this is true? I though that the comparison tests says that if $\lim_{n\to\infty} \frac{a_n}{b_n}\to L$ where $0<L<\infty$, then $a_n$ and $b_n$ follow the same convergence. So wouldn't this mean that $\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$ diverges since $\frac{1}{n}$ diverges?

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  • $\begingroup$ I think you have confused convergence with divergence. $\endgroup$ – Clayton Oct 26 '20 at 23:17
  • $\begingroup$ @Clayton You mean it actually diverges and my thinking is correct? $\endgroup$ – mathim1881 Oct 26 '20 at 23:19
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    $\begingroup$ Yes, the series is divergent and you are thinking along the right lines. $\endgroup$ – Kavi Rama Murthy Oct 26 '20 at 23:20
  • $\begingroup$ @Clayton Or the professor did. $\endgroup$ – zhw. Oct 27 '20 at 1:12
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What we are using is limit comparison test and since

$$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}=\frac{\sin{\frac{2}{\sqrt{n}}}}{\frac{2}{\sqrt{n}}}\frac{2}{3+2\frac{\sqrt n}n} \to \frac23$$

as you noticed the series diverges.

Note that limit comparison test also works for the two limiting cases

  • $b_n$ converges and $\frac{a_n}{b_n} \to 0 \implies a_n$ converges
  • $b_n$ diverges and $\frac{a_n}{b_n} \to \infty \implies a_n$ diverges

As an alternative by direct comparison test, using that $\sin x\ge x-\frac16 x^3$ we have

$$\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2} \ge \frac{\frac{2}{\sqrt{n}}-\frac{4}{3n\sqrt{n}}}{3\sqrt{n}+2}=\frac{2-\frac{4}{3n}}{3n+2\sqrt{n}}\ge \frac{\frac23}{3n+3n}=\frac19\frac1n$$

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  • $\begingroup$ Thanks, I appreciate this concise and complete explanation! $\endgroup$ – mathim1881 Oct 26 '20 at 23:40
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    $\begingroup$ @mathim1881 You are welcome! The limiting cases are not always clearly stated. Morover the example by direct comparison test shows how much LCT is effective to determine the behaviour of the series. Bye $\endgroup$ – user Oct 26 '20 at 23:42
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Yes, the series in question does diverge. What your professor used is known as the Limit Comparison Test. The proof of the test can be found in the Wikipedia article.

A non-rigorous, intuitive way to think about this is that if one series diverges to infinity, while the other converges to a real positive number, then you know that taking the limit of $\frac{a_n}{b_n}$ will result in either $\infty$ in the numerator or in the denominator and thus a limit of either $0$ or $\infty$.

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  • $\begingroup$ Thanks very much, the intuitive explanation definitely helped me confirm some thoughts I had. $\endgroup$ – mathim1881 Oct 26 '20 at 23:39
  • $\begingroup$ No problem! Do you have any additional questions? $\endgroup$ – General Poxter Oct 27 '20 at 0:00

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