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$$\sum_{n=1}^{\infty} \sin{(\sqrt{n+1}-\sqrt{n})}$$

Here are my ideas:

If we prove $\lim_{n\to \infty}a_n$ isn't $0$, series diverge. Dunno how to prove it though.

Use $\lim_{x\to 0}\frac{\sin{x}}{x}=1$ .

Can someone give me a hint about solution?

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    $\begingroup$ Actually since $\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$, you have $\lim_{n \rightarrow 0} \sin(\sqrt{n+1} - \sqrt{n}) = 0$ $\endgroup$
    – iamvegan
    Sep 2, 2016 at 11:15

2 Answers 2

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Hint: for large $n$ we have $\frac{1}{4\sqrt{n+1}} < \sin(\sqrt{n+1} - \sqrt{n})$

Then $$ \sum_N^\infty \sin(\sqrt{n+1} - \sqrt{n}) > \sum_N^\infty \frac{1}{4\sqrt{n+1}} $$

which is divergent.

Explanation of the Hint: Since $\lim_{x \rightarrow 0} \dfrac{\sin x}{ x} = 1$. When $x$ is sufficiently small we have $$\dfrac{3}{2} \geq|\dfrac{\sin x}{x}| \geq \dfrac{1}{2}$$ This implies $$ \frac{1}{4\sqrt{n+1}} < \frac{\sqrt{n+1} - \sqrt{n}}{2} < \sin(\sqrt{n+1} - \sqrt{n}) $$

Note: Sorry for the change from $\frac1{2}$ to $\frac1{4}$ in the hint.

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    $\begingroup$ Isn't $\sin {x} \lt x $ for $x\in \langle 0,1]$, so how do you exactly prove this : $\frac{1}{2\sqrt{n+1}} < \sin(\sqrt{n+1} - \sqrt{n})$ $\endgroup$ Sep 2, 2016 at 11:27
  • $\begingroup$ @RushThaMan I included the explanation. $\endgroup$
    – iamvegan
    Sep 2, 2016 at 12:39
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Let $a_n = \sqrt {n+1} - \sqrt n.$ Then $a_n = 1/(\sqrt {n+1} + \sqrt n).$ Therefore $a_n\to 0^+$ and we see $(\sin (a_n))/a_n \to 1.$ So by the limit comparison test, $\sum \sin(a_n)$ converges iff $\sum a_n$ converges. Since $a_n > 1/(2\sqrt {n+1}),$ $\sum a_n$ diverges and we're done.

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