In the following questions I am trying to find a function with the given properties or explain why no such function exists.
1) An infinitely differentiable function on $R$ with the Taylor series that only converges on $(-1,1)$
For this I have chosen the geometric series,
$$\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=0}^{\infty}x^n$$
So this series is a geometric series and only converges when $|r|<1$. Thus, converges only for $(-1,1)$ but is this series infinitely differentiable?
2) An inifinitely differentiable function on $R$ with a Taylor Series that only converges for $x\le0$
Would the following series work? $$\sum _{n=0}^{\infty}\frac{(-x)^{n}}{n}$$
No. That function fails to be differentiable at $x = 1$. But you're on the right track. The radius of convergence of the power series is always less than the distance to the nearest singularity. But to be differentiable on $\mathbb R$, it can't have any singularities for $x \in \mathbb R$. How do we reconcile this?
No. Aside from the fact that, as written, the first term is 1/0, if we start from $n=1$ that is the series for $-\ln|1+x|$, which converges for $|x|<1$ and is not everywhere differentiable. The answer to this question depends on whether we require the power series to be about 0 and whether we need it to converge for all $x \le 0$ or just for no values $x > 0$. It's also worth noting that there exist infinitely differentiable functions with a convergent power series at $x = 0$, but that series only converges to the value of the function for $ x \le 0$.