Vtyjkyuk

We define $f_n(x) = fracn x^2 sin(nx)n^4 + x^4$. We have to prove that $$ sum_n=1^infty f_n $$ converges uniformly to a derivative function on $mathbbR$. To prove that, it suffices to show that there exists $x_0 in mathbbR$ so that $sum_n=1^infty f_n(x_0)$ converges (just take $x_0 = 0$) and that $$ sum_n=1^infty f_n'$$ converges uniformly to a certain function on $mathbbR$. If we do the derivative, we obtain $$ f_n'(x) = frac(2nx sin(nx) + n^2x^2 cos(nx))(n^4+x^4) - 4nx^5 sin(nx)(n^4+x^4)^2. $$ Now, to show the uniform convergence of the series, I think we must use the Weierstrass criterion, but I don't know how to obtain an upper bound valid on $mathbbR$ for $f_n'$.

Can anyone help me? Thank you.

1) Uniform convergence of $sum_n=1^infty f_n$ in $mathbbR$: note that
$$sum_n=1^infty f_n(x)=x^2sum_n=1^infty fracn sin(nx)n^4 + x^4$$
and $left|fracnsin(nx)n^4 + x^4right|leq frac1n^3$, then by the $M$-test the series converges uniformly in $mathbbR$ to some continuous function $F$.

2) As regards the differentiability of $F$, all you need is the convergence of $sum_n=1^infty f_n'$ in the compact set $[-R,R]$ for all $R>0$: for $xin [-R,R]$, and for $ngeq1$,
$$|f'_n(x)|leq frac(2nR + n^2R^2)(n^4+R^4) + 4nR^5n^8leq fracC_Rn^2$$
for some constant $C_R$ (which depends on $R$). Then $sum_n=1^infty f_n$ converges uniformly to the derivative of $F$ on $[-R,R]$ for all $R>|x_0|$ (recall that differentiability is a local property).