Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was thinking about derivative of infinite sum of functions, i.e.

$$f(x) = \sum_{i = 0}^\infty g_i(x)$$

$g(x)$ is continuous in domain of $f$

Because if $(f+g)'(x) = f'(x) + g'(x)$ then $\left(\sum\limits_{i = 0}^{\infty} g_i(x)\right)' = \sum\limits_{i = 0}^{\infty} g_i'(x)$ isn't it?

share|cite|improve this question
Your function $f$ is not well-defined except $g(x) = 0$ for all $x$. I suppose you mean something different. Also remember that having $g$ being continuous does not make it differentiable. – Matthias Klupsch Nov 24 '11 at 18:33
That doesn't always work. The canonical example is due to Weierstrass... – J. M. Nov 24 '11 at 18:35
OK, but what if I set $g(x) = \frac{1}{n^x}$. In this question I'm interested especially in $\zeta(z)$ function – Łukasz Niemier Nov 24 '11 at 18:37
In that case... – J. M. Nov 24 '11 at 18:43
up vote 7 down vote accepted

First I assume you mean $g_i$ instead of $g$, and you have to suppose at least that the $g_i$ are all differentiable (more than just continuous).

Even then this is in general false. One common case where it is true is when you assume uniform convergence of $\sum g_i^{'}$ and at least one point of convergence for $\sum g_i$.

A counter example under your hypothesis : take $g_i^{'}(x) = \cos(i \pi x)/i^2$. then $\sum g_i$ converges since it converges normally ($\sum \frac{1}{i^2}< \infty$) but $\sum g_i^{'}$ diverges at 0 (since $\sum \frac{1}{i} = \infty$).

share|cite|improve this answer
And what if $g_i(x) = 1/i^x$? That is especially interesting me. – Łukasz Niemier Nov 24 '11 at 18:46
then it is true for $x>1$ because the convergence is uniform on all compacts of $]1, +\infty[$ – Glougloubarbaki Nov 24 '11 at 18:53
$\sum_{i=1}^\infty 1/i^x$ is the Riemann zeta function $\zeta(x)$ for $\Re x > 1$. The series of derivatives $\sum_{i=1}^\infty -\ln(i)/i^x$ also converges for $\Re x > 1$, and uniformly on compact sets, so by the "One common case" Glougloubarbaki mentioned the sum is indeed $\zeta'(x)$ for $\Re x > 1$. – Robert Israel Nov 24 '11 at 18:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.