There is a function:
$V(n) = \sum_{i=1}^\infty f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)$,
where $f$,$b$ and $a$ are $C^\infty$ smooth functions.
I know that we can prove that the function $V(n)$ is also smooth under some assumptions.
I SUGGEST THE NEXT SOLUTION:
We know that if
$\forall i$ $ f_i() \in C^{1} $
$s(n)=\sum_{i=1}^\infty f_i(n)< \infty$ (not neceesary uniformly)
and
$\sum_{i=1}^\infty f_i^{'}(n)=\sigma(n)$ uniformly
then:
$s(n) \in C^{1}$ and $s^{'}(n)=\sigma(n)$
So IN MY CASE we have $C^{\infty} $
$s(n)=\sum_{i=1}^\infty f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)< \infty$
and we need to show that:
$\sum_{i=1}^\infty h_i^{'}(n)$ converges uniformly,
where $h_i(n)=f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)$
MY QUESTION: Can somebody help me with any ideas how I can prove that $\sum_{i=1}^\infty h_i^{'}(n)$ converges uniformly.
I will really appreciate your help!
If we don't place any further restrictions on $f,b_i$ and $a_i$, then it's not true that $\sum h_i(n)$ converges uniformly.
Let $\{U_i\}$ be a partition of unity on $\Bbb R$, and let $H_i$ be an antiderivative of $U_i$. We want to show that there exist smooth functions $f, b_i$ and $a_i$ such that $H_i(n)=f(b_i(n))b_i^{'}(n)- f(a_i(n))a_i^{'}(n)$. This is easily achieved: take $f\equiv 1$ and $a_i\equiv 0$ to be constant functions, and $b_i$ an antiderivative of $H_i$.
By definition, $\displaystyle\sum_{i=1}^\infty H_i(n) = \sum_{i=1}^\infty U_i(n) = 1$ for all $n$. However, this convergence cannot be uniform because the support of each $U_i$ is compact, hence the support of any partial sum is compact. In particular, for any $K$ there is always some number $n_0$ such that $\displaystyle \sum_{i=1}^K U_i(n_0) = 0$, and this is automatically an obstruction to uniform convergence.