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:

$\underset{n\to \mathrm{\infty}}{lim}(\frac{\mathrm{sin}(2\sqrt{1})}{n\sqrt{1}\mathrm{cos}\sqrt{1}}+\cdots +\frac{\mathrm{sin}(2\sqrt{n})}{n\sqrt{n}\mathrm{cos}\sqrt{n}})=\underset{n\to \mathrm{\infty}}{lim}(\frac{2\mathrm{sin}(\sqrt{1})}{n\sqrt{1}}+\cdots +\frac{2\mathrm{sin}(\sqrt{n})}{n\sqrt{n}})$

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:

$\frac{2\mathrm{sin}(\sqrt{1})}{n\sqrt{1}}+\cdots +\frac{2\mathrm{sin}(\sqrt{n})}{n\sqrt{n}}\le \frac{2}{n\sqrt{1}}+\cdots +\frac{2}{n\sqrt{n}}\le \frac{2}{n}\cdot n\u27f62$

What am I missing?

$\underset{n\to \mathrm{\infty}}{lim}(\frac{\mathrm{sin}(2\sqrt{1})}{n\sqrt{1}\mathrm{cos}\sqrt{1}}+\cdots +\frac{\mathrm{sin}(2\sqrt{n})}{n\sqrt{n}\mathrm{cos}\sqrt{n}})=\underset{n\to \mathrm{\infty}}{lim}(\frac{2\mathrm{sin}(\sqrt{1})}{n\sqrt{1}}+\cdots +\frac{2\mathrm{sin}(\sqrt{n})}{n\sqrt{n}})$

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:

$\frac{2\mathrm{sin}(\sqrt{1})}{n\sqrt{1}}+\cdots +\frac{2\mathrm{sin}(\sqrt{n})}{n\sqrt{n}}\le \frac{2}{n\sqrt{1}}+\cdots +\frac{2}{n\sqrt{n}}\le \frac{2}{n}\cdot n\u27f62$

What am I missing?

Brendon Fernandez

Beginner2022-06-20Added 14 answers

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

$|\frac{2\mathrm{sin}(\sqrt{1})}{n\sqrt{1}}+...+\frac{2\mathrm{sin}(\sqrt{n})}{n\sqrt{n}}|\le 4\frac{1}{\sqrt{n}}\to 0.$

Note that $\mathrm{sin}(\cdot )$ can change signs, so absolute value is needed in the estimate.

$\sum _{k=1}^{n}\frac{1}{\sqrt{k}}\le 1+{\int}_{1}^{n}\frac{1}{\sqrt{x}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le 2\sqrt{n},$

yields

$|\frac{2\mathrm{sin}(\sqrt{1})}{n\sqrt{1}}+...+\frac{2\mathrm{sin}(\sqrt{n})}{n\sqrt{n}}|\le 4\frac{1}{\sqrt{n}}\to 0.$

Note that $\mathrm{sin}(\cdot )$ can change signs, so absolute value is needed in the estimate.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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