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So I have to assess the convergence of $$\displaystyle\sum_{n=1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right).$$

I'm told that it diverges, but can't really see why.

The divergence test doesn't really help, because $\lim\limits_{x\to\infty}\displaystyle\frac{1}{\sqrt{n}}=0$, so

$\lim\limits_{x\to\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right)=0$, which doesn't conclude its divergence.

I doubt the ratio test would be much of use in this situation.

I can't imagine using the integral comparison test, as I wouldn't know where to start with $\displaystyle\int_{1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right) \mathrm dx$.

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Limit-compare to $\sum_n \frac{1}{\sqrt{n}}$. You need the limit $$ \lim_{n \to \infty} \frac{\sin(1/\sqrt{n})}{1/\sqrt{n}} $$ which is $1$ by a change of variables, making use of $$ \lim_{t \to 0^+} \frac{\sin t}{t}=1. $$ To get from point A to point B, change variables by $t=\frac{1}{\sqrt{n}}$. As $n \to \infty$, $t \to 0^+$.

Since $\sum_n \frac{1}{\sqrt{n}}$ diverges, so does your series.

The terms in your series are positive, so this is legal.

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  • $\begingroup$ Ok, I understand it now. I would, to improve the clarity of your answer for future readers, elaborate on the change of variables explanation, i.e. how you got from $\lim\limits_{n \to \infty} \displaystyle \frac{\sin(1/\sqrt{n})}{1/\sqrt{n}}$ to $\lim\limits_{t \to 0} \displaystyle \frac{\sin t}{t}=1.$ $\endgroup$ – user98937 Dec 1 '17 at 13:01
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Use the inequality that $\sin x\geq\dfrac{2}{\pi}x$ for $x\in[0,\pi/2]$.

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  • $\begingroup$ Could you explain how you arrive at the comparison please? $\endgroup$ – user98937 Dec 1 '17 at 13:03
  • $\begingroup$ The inequality comes from the convexity of $\sin$ on $[0, \pi/2]$. $\endgroup$ – B. Mehta Dec 1 '17 at 14:13
  • $\begingroup$ Perhaps you can look at the derivative of the function $\dfrac{\sin x}{x}$ on $(0,\pi/2]$ and try to argue that it is decreasing. $\endgroup$ – user284331 Dec 1 '17 at 17:10
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You can do a limit comparison with $\sum_{n\geq1}\frac{1}{\sqrt{n}}$, since

$$ \lim_{n\to\infty}\sqrt{n}sin(\frac{1}{\sqrt{n}}) = 1 $$

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The function $\sin(x) \geq\dfrac{2}{\pi}x$ for $x\in[0,\pi/2]$

$\sum_{n} 1/n^{1/2}$ diverge

$\forall n: \sin(1/n^{1/2}) \geq \dfrac{2}{\pi n^{1/2}}$

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