# Series of functions: convergence, continuity, differentiability

I'm trying to work on the function series problems and I'm confused with the necessary and sufficient conditions for a function $$f(x)=\sum_{n=1}^{\infty} f_n(x)$$ to be:

1. well-defined
2. continuous
3. differentiable

and how these conditions are connected with pointwise / uniform convergence of the series. My thoughts:

• pointwise convergence implies $(1)$
• uniform convergence of the series of functions and the series of the derivatives implies $(3)$

But I'm not sure how to deal with $(2)$. For example, if I know that $f(x)$ is well-defined and I try to use the definition of continuous function: $$\forall \epsilon >0 \text{ and } \forall x \exists \delta = \delta(x,\epsilon) \text{ s.t. } |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$$

what should I know about the series to write the condition in the following way: $$|f(x)-f(y)|=|\sum_{n=1}^{\infty}\left( f_n(x)-f_n(y)\right)|$$ Is the pointwise convergence enough? Or I need the series to converge uniformly?

I would be thankful if you help me to understand these concepts on the following example:

Example: Find all x for which the series satisfies conditions $(1)-(3)$: $$f(x)=\sum_{n=1}^{\infty} \frac{n^x}{2^n}$$

So, by the ratio test for each fixed x the series is convergent, therefore $f(x)$ is well defined $\forall x \in \mathbb{R}$: $$\lim_{n \to \infty}\left|\frac{(n+1)^x2^n}{n^x2^{(n+1)}}\right|=\lim_{n \to \infty}\left|\frac{1}{2}(1+\frac{1}{n})^x\right| < 1$$

Is it true that using Dirichlet test I can show that the series converges uniformly on any compact set? If it is not true, how should I approach this? And how should I deal with unbounded intervals, $(-\infty,a]$ and $[a,\infty)$? Then as I get the intervals of pointwise/uniform convergence, how should I get continuity and differentiability?

• For continuity of the sum, pointwise convergence is not enough; uniform convergence on compact sets is enough (provided that the functions $f_n$ are continuous), but not necessary. For differentiability, convergence at one point plus uniform convergence of the series of derivatives on compact sets is enough. In your example, you do have uniform convergence on compact sets, and also uniform convergence of the series of derivatives on compact sets, because if $K$ is a compact subset of $\mathbb R$, say $K\subset [-a,a]$, then $n^x\leq n^a$ and $\vert xn^{x-1}\vert\leq an^{a-1}$ on $K$. – Etienne Feb 13 '14 at 23:00