If `|a_1sinx+a_2sin2x++a_nsinn x|lt=|sinx|` for `x in R ,` then prove that `|a_1+2a_2+3a_3+n a_n|lt=1`

Explanation

Clearly, we can get `a_(1)+2a_(2)+3a_(3)+…+na_(n),` by differentiating `a_(1) sin x +a_(2) sin 2x + …+a_(n)` sin nx and then putting x=0.
Thus, we have to prove that `|f'(0)|le1.`
where, `f(x)=a_(1) sin x +a_(2) sin 2x +…+a_(n) sin nx`
`therefore" "f'(x)=a_(1) cos x +2a_(2) cos 2x +...+na_(n) cos nx`
`"or "f'(0)=a_(1)+2a_(2)+...+na_(n)`
Also, given `|f(x)|le| sin x |" for "x in R" (1)"`
Put x=0. Then `|f(0)|le or f(0)=0.` Now,
`f'(0)=underset(hrarr0)lim(f(h)-f(0))/(h)=underset(hrarr0)lim(f(h))/(h)" [as f(0) = 0]"`
`"or "|f'(0)|=underset(hrarr0)lim|(f(h))/(h)|leunderset(hrarr0)lim|( sin h)/h|=1 [as |f(x)|le| sin x|]`
`"Hence, "|f'(0)|le1.`