Termwise differentiation for absolutely convergent series Suppose $f(x,y)=\sum_{n=1}^\infty f_n(x,y)$ converges absolutely. Is it true that $$\dfrac{\partial}{\partial x}f(x,y)=\sum_{n=1}^\infty\dfrac{\partial}{\partial x}f_n(x,y)?$$
If not, what extra condition do we need (e.g. uniform convergence)?
 A: No, that does not work already for functions of one variable. The theorem you want to use is:
Theorem. Let $I$ be a nonempty open interval in $\mathbb{R}$. Let $(f_n)$ be a sequence of ${\cal C}^1$ functions on $I$ such that


*

*$(f_n)$ converges to some function $f$ pointwise

*$(f_n)'$ converges uniformly to some function $g$ (locally uniformly is enough)


Then $f$ is differentiable and $f' = g$.
I trust you will know how to adapt this to series? The theorem is also true for functions of several variables, then the uniform convergence of $f'$ (which is a differential now) is taken with respect to a subordinate (matrix) norm.
For a counter-example of this theorem when only assumes the uniform convergence of $f_n$, consider $f_n(x) = \sqrt{x^2 + 1/n}$. I think this also provides a counter-example for the series version of the theorem with only normal convergence of the series (check that $\sum (f_{n} - f_{n+1})$ converges normally).
A: Absolute convergence isn't relevant to the identity in question. You need something stronger than uniform convergence: see http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability for example. (I think the presence of the variable $y$ is a red herring here.)
