Can You Prove the Inequality in This Week's Math Challenge?

[anemone]

Here is this week New Year's POTW::)

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Let $g(x)=a_1\sin x+a_2\sin 2x+\cdots+a_n\sin nx$, where $a_1,\,a_2,\,\cdots,\,a_n$ are real numbers. Suppose that $|g(x)|<|\sin x|$ for all real $x$.

Prove $|a_1+2a_2+\cdots+na_n|<1$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to vidyarth for his correct solution.:)

You can find the suggested solution below:

Suggested solution:

Spoiler

Observe that $g(0)=0$ and $g'(0)=a_1+2a_2+\cdots+na_n$.

On the other hand,

\(\displaystyle \begin{align*}\left| g'(0) \right|&=\left|\lim_{{x}\to{0}}\frac{g(x)-g(0)}{x-0} \right|\\&=\left|\lim_{{x}\to{0}}\frac{g(x)}{x} \right|\\&<\left|\lim_{{x}\to{0}}\frac{\sin x}{x} \right|\\&=1\,\,\text{Q.E.D.}\end{align*}\)