taghdh9

2022-06-19

Using the trigonometric identity of $\mathrm{sin}2\alpha =2\mathrm{sin}\alpha \mathrm{cos}\alpha$, I rewrote the expression to:
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{sin}\left(2\sqrt{1}\right)}{n\sqrt{1}\mathrm{cos}\sqrt{1}}+\cdots +\frac{\mathrm{sin}\left(2\sqrt{n}\right)}{n\sqrt{n}\mathrm{cos}\sqrt{n}}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{2\mathrm{sin}\left(\sqrt{1}\right)}{n\sqrt{1}}+\cdots +\frac{2\mathrm{sin}\left(\sqrt{n}\right)}{n\sqrt{n}}\right)$
I then tried using the squeeze theorem to get the limit, but I can't get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two:
$\frac{2\mathrm{sin}\left(\sqrt{1}\right)}{n\sqrt{1}}+\cdots +\frac{2\mathrm{sin}\left(\sqrt{n}\right)}{n\sqrt{n}}\le \frac{2}{n\sqrt{1}}+\cdots +\frac{2}{n\sqrt{n}}\le \frac{2}{n}\cdot n⟶2$
What am I missing?

Brendon Fernandez

The inequality
$\sum _{k=1}^{n}\frac{1}{\sqrt{k}}\le 1+{\int }_{1}^{n}\frac{1}{\sqrt{x}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le 2\sqrt{n},$
yields
$|\frac{2\mathrm{sin}\left(\sqrt{1}\right)}{n\sqrt{1}}+...+\frac{2\mathrm{sin}\left(\sqrt{n}\right)}{n\sqrt{n}}|\le 4\frac{1}{\sqrt{n}}\to 0.$
Note that $\mathrm{sin}\left(\cdot \right)$ can change signs, so absolute value is needed in the estimate.

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