If we have $\sum_{i=1}^\infty f_i(x)$ and assume this is a convergent sum and asumme all the $f_i$ are differentiable in every point. Is the derivative of the infinite sum equal to the sum of the derivatives (so is $(\sum_{i=1}^\infty f_i(x))' = \sum_{i=1}^\infty f'_i(x)$ ) ?
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$\begingroup$ I don't believe that is always true, you need to put some conditions on the sequence $f_i$ to be able to do that. $\endgroup$– Gregory GrantMay 23, 2015 at 13:09
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$\begingroup$ See here: math.stackexchange.com/questions/147869/… $\endgroup$– Gregory GrantMay 23, 2015 at 13:12
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1 Answer
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This is called "differentiation term-by-term". You can come up with examples where it fails. But in many cases it is correct. The best known case is the power series. In that case, differentiation term-by-term holds inside the radius of convergence.
Another one is due to Fubini ... if all $f_i$ are increasing, then term-by-term differentiation holds almost everywhere.
