$\sum_1^\infty e^{n(1-x)}x^n$ and it's Uniform/Pointwise Convergence on different intervals

Question

Find largest subset $X \subset [0,\infty)$ on which $\sum_1^\infty e^{n(1-x)}x^n$ is convergent:

Find the region where $\sum_1^\infty e^{n(1-x)}x^n<\infty$.

Find $n+1$ term and the $n$ terms. $\lim_{n\rightarrow \infty} \frac{e^{(n+1)(1-x)}x^{n+1}}{e^{n(1-x)}x^n}=e^{1-x}x$

The radius of convergence is $x\in X$ s.t. $e^{1-x}x<1$ which I think is $0\leq x < 1 $ and $1<x<\infty$. (I'm not sure if that is correct).

Next Part: Prove that $\sum_1^\infty e^{n(1-x)}x^n$ is uniformly convergent on $[0,a]$ and $[b,\infty)$ where $a<1$ and $b>1$. Is it uniformly convergent on $X$?

For this mark I am thinking of using the M test. Clearly the largest values are obrained when $x=a$ or $x=b$, thus take $M=\max {f(a),f(b)}$ (sorry for bad notation) which is $<1$. Thus since $\sum M $ is geometric series $\rightarrow$ it's $<\infty$. Thus it's uniform.

Can I simply state that is is also uniform on $X$ by taking $a\rightarrow 1$ and $b\rightarrow 1$?

Answer 1

For the first part, note that the series is not a power series so you can't talk about the "radius of convergence". Note that your series is a geometric series in disguise:

$$ \sum_{n=1}^{\infty} e^{n(1-x)} x^n = \sum_{n=1}^{\infty} \left( e^{1 - x} x \right)^n = \sum_{n=1}^{\infty} y^n $$

where $y = y(x) = e^{1-x} x$. Thus, the series converges when $y < 1$ and diverges when $y \geq 1$ which implies that the original series converges on $X := [0, 1) \cup (1, \infty)$ and diverges when $x = 1$.

On $X$, the series converges to the function

$$ f(x) := \frac{y(x)}{1-y(x)} = \frac{e^{1-x}x}{1 - e^{1-x}x}. $$

The function $f$ is not bounded on $X$ (as $\lim_{x \to 1} f(x) = \lim_{y \to 1^{-}} \frac{y}{1-y} = +\infty$) and thus the series cannot converge uniformly on $X$ because a uniform limit of bounded functions must be bounded.

Finally, if $x \in [0,a] \cup [b, \infty)$ where $0 < a < 1 < b$ then set $M = \max \{y(a), y(b) \}$. We have $0 \leq y(x) \leq M$ and $0 < M < 1$. Since the geometric series $\sum_{n = 1}^\infty M^n$ converges, the Weirstrass $M$-test implies that $\sum_{n=1}^{\infty} y(x)^n = \sum_{n=1}^{\infty} e^{n(1-x)}x^n$ converges uniformly on $[0,a] \cup [b, \infty)$.

Answer 2

As said, the series converges for all x in $X=[0,\infty]-\{1\}$. Its justification is that in $x=1$ occurs the only maximum value of $g(x)=xe^{1-x}$, being this 1(check it deriving). Note too that $g$ is non-negative in $[0,\infty]$. Putting all this together we have that $|g(x)|<1$ when x is non-negative and different from 1, and so the geometric series $g^n$ converges for those values. Second part: As g is "decreasing" to the right and left of $x=1$ we got that $g^n$ is also decreasing in that way.Hence: $|g^n(x)|\le g^n(a)$ with $g(a)<1$ forall $x$ in $[0,a]$ and all $n$ in $\mathbb{N}$ and by the M test it converges uniformly in $[0,a]$. Analogous in $[b,\infty]$. It can't converge uniformly on $X$, because this would imply that it also converges in $1$, which is not true. See this Uniformly convergent in a set implies uniformly convergent in the set closure, too.