# Uniform convergence of series with utility function

Let $u(x)$ be a real function with the properties

• $u(x)$ is continuous and non-decreasing in $x$
• $u'(x)$ is non-increasing in $x$
• $u(0)=0$
• $u'(0)=1$.

In other words, $u(x)$ is a utility function.

In a rather old insurance paper the series on the left hand side of $$\sum_{k=0}^{\infty} \frac{\lambda^k}{k!}u(-\lambda u(-z)-zk)=0,$$ where $z>0$ and $\lambda>0$, is differentiated with respect to $\lambda$ rather non-chalantly and without further comment by differentiation of every summand, i.e differentiation and summation are exchanged. Is the uniform convergence of the series consisting of the derivatives of the summands, i.e $$\sum_{k=0}^{\infty} \left(\frac{\lambda^{k-1}}{(k-1)!}u(-\lambda u(-z)-zk)-\frac{\lambda^k}{k!}u'(-\lambda u(-z)-zk)u(-z)\right),$$ needed for this exchange to work really that easy to see? To be more specific, how would such a general series be approached?

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 differentiation is a linear operator and summation as well...as I see it, you can exchange it with no problem. – Cristian Sep 18 '12 at 17:09 @Cristian Not true, Weierstrass M-test or showing series is cauchy is only thing that comes to mind. But I don't see it working unless $\lambda$ has some constraint. – Stuart Sep 18 '12 at 17:12 @user19192 I missed the constraint $\lambda>0$ when writing the question, but I guess this is not sufficient. It is interesting that the step is, in a sense, correct, as the equations resulting from this differentiation can be derived via another way as well. – Mycroft Sep 18 '12 at 17:19 @Mycroft does your original eq holds as an identity for $\lambda>0$? – Cristian Sep 18 '12 at 17:23 @Cristian It does. – Mycroft Sep 18 '12 at 17:35