Can the Sum Rule for derivatives be extended to infinite series?

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?

• Yes, it works beautifully except when it doesn't. Sep 5, 2012 at 1:48

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)

• 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.) 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 $(-1, 1)$.

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