I wrote an answer here, which I'm not sure works.

The sum rule for differentiation of two functions says that $D(u+v) = D(u) + D(v)$ where $D$ indicates the derivative, and $u$ and $v$ two functions. The sum rule can get extended to any finite set of functions. Since numbers can get regarded as functions, this implies that for any finite series $S=a + b + \dots+z$ we can evaluate $D(S).$ Can we extend the sum rule to differentiation of convergent infinite series? Divergent infinite series? Why or why not?

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    $\begingroup$ Yes, it works beautifully except when it doesn't. $\endgroup$ Sep 5, 2012 at 1:48

2 Answers 2


Not really. Actually, what you want is uniform convergence and majorant series.

DEFINITION 1 Let $f_n(x)$ be a sequence of functions. In particular, suppose $f_n(x)=\sum_{k=0}^n g_k(x)$ for some sequence $\{g_k\}_{k\in \mathbb N}$ of functions. Let $D$ be the set of points $x$ such that $\lim f_n(x)$ exists. Call $D$ the domain of convergence of $f=\lim f_n$.

An important property is a series might have is being majorant.

DEFINITION 2 We say that a series of functions is majorant in a certain domain $D'$ if there exists a convergent positive series $A=\sum a_k$ such that, for each $x$ in that domain $D'$ we have $|g_k(x)|\leq a_k$. Given a series $f=\lim f_n=\lim\sum^n g_k$, we say that $f$ converges absolutely if $f^*=\lim\sum^n |g_k|$ converges. (Thus, a majorant series is absolutely convergent.)

Yet another important case scenario is uniform convergence:

DEFINITION 3 (Uniform convergence) We say a series of functions converges uniformly in $D$ if for all $\epsilon>0$ there is an $N$ (depending only on $\epsilon$), such that $n\geq N$ implies $$|f(x)-f_n(x)|<\epsilon $$

We usually say $N$ is independent of the choice of $x$, too. You can picture this behaviour as follows: Each partial sum is always contained in the strip inside $f(x)+\epsilon$ and $f(x)-\epsilon$ of width $2\epsilon$.

In particular, every majorant series converges uniformly. This is known as Weierstrass' $M$ criterion. For majorant series, the following is valid:

THEOREM 1 If the series $\sum u_k(x)$ composed of functions with continuous derivates on $[a,b]$ converges to a sum function $s(x)$ and the series $$\sum u'_k(x)$$ composed of this derivatives is majorant on $[a,b]$, then $$s'(x)=\sum u'_k(x)$$

This stems from

THEOREM 2 Let $s(x)=\sum u_k(x)$ be a series of continuous functions, majorant on some $D$. Then, if $x$ and $\alpha$ are in $D$

$$\int_\alpha^x s(t)dt=\sum\int_\alpha^xu_k(t)dt$$

You can read this in much more detail, and find proofs, in (IIRC) Apostol's Calculus (Vol.1)

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    $\begingroup$ For those interesed in a more detailed analysis of a closely related issue, see my 29 December 2006 sci.math post. Incidentally, at the end of that post I describe a possible research topic. (I mention this since least once in StackExchange someone has asked for an example of a research topic in real analysis.) $\endgroup$ Sep 5, 2012 at 15:44

The sum rule is fine for absolutely convergent infinite series. For conditional convergence (i.e. at the boundary of the interval of convergence) you will run into problems. For instance $$ \frac{1}{1-x} = 1 + x + x^2 + \ldots $$ is conditionally convergent at $x = -1$, so the interval of convergence is $[-1, 1)$; taking derivatives, we get $$ \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \ldots $$ and now the interval of convergence is just $(-1, 1)$.

  • $\begingroup$ The derivative of $1/(1-x)$ is $1/(1-x)^2$. $\endgroup$ Sep 5, 2012 at 0:14
  • $\begingroup$ devastating :-) correcting the text (I had a log before the correction!) $\endgroup$
    – user29743
    Sep 5, 2012 at 1:37
  • $\begingroup$ This does not seem to work for the second derivative $-2/(1-x)^3$ is not $2+6x+12x^2+...$ but rather the negative of it. What is the reason for this? $\endgroup$ Dec 3, 2015 at 7:40
  • $\begingroup$ According to Wikipedia "Sum rule in differentiation": "Note this does not automatically extend to infinite sums." That answers my question. $\endgroup$ Dec 3, 2015 at 7:47
  • $\begingroup$ @DanielCentore, The derivative of $\dfrac1{(1-x)^2}$ is not $-\dfrac2{(1-x)^3}$, but $\dfrac2{(1-x)^3}$ $\endgroup$
    – user228113
    Dec 28, 2015 at 18:29

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