ps: Due to my poor English, I might describe my thought roughly.
Suppose $\{a_n\} (n \ge 0)$ is a sequence consisting of non-negative numbers, and $\{a_n\}$ satisfies that forall $x > 0$, $f(x) = \sum_{n\geq 0} e^{-nx} a_n$ is convergent. Calculate $\lim_{x \rightarrow 0^{+}} x f(x)$.
My thought:
Let $S = \sum_{n = 0}^{\infty} a_n (e^{-x}) ^{n}$
then $e^{-x}S = \sum_{n = 1}^{\infty} a_{n-1} (e^{-x})^n$
so $S = \frac{1}{1-e^{-x}}\sum_{n = 1}^{\infty}(a_n - a_{n-1})(e^{-x})^n + a_0$
so $\lim_{x \rightarrow 0^{+}} \frac{x}{1-e^{-x}}\sum_{n = 1}^{\infty}(a_n - a_{n-1})(e^{-x})^n + a_0 = \sum_{n = 1}^{\infty}(a_n - a_{n-1})(e^{-x})^n + a_0$
and then I have been discouraged
can anyone give me some hints or the true answer?
any hints welcome.
Assume the sequence $a_n$ is convergent to $a.$ Then $$xf(x)=x\sum_{n=0}^\infty e^{-nx}a_n=x\sum_{n=0}^\infty e^{-nx}[a_n-a]+a{x\over 1-e^{-x}}$$ The second term tends to $a.$ We will show that the first term tends to $0.$ Indeed, for a fixed $N$ we have $$g(x):=x\left |\sum_{n=0}^\infty e^{-nx}[a_n-a]\right|\le x\sum_{n=0}^\infty e^{-nx}|a_n-a|\\ =x\sum_{n=0}^{N-1} e^{-nx}|a_n-a|+x\sum_{n=N}^\infty e^{-nx}|a_n-a|\\ \le x\sum_{n=0}^{N-1} e^{-nx}|a_n-a|+\left (\sup_{n\ge N}|a_n-a|\right )\,x\sum_{n=N}^\infty e^{-nx} \\ =x\sum_{n=0}^{N-1} e^{-nx}|a_n-a|+\left (\sup_{n\ge N}|a_n-a|\right )\,{x\over 1-e^{-x}}e^{-Nx}$$ Taking the limit $x\to 0^+$ gives $$\limsup_{x\to 0^+}g(x)\le \sup_{n\ge N}|a_n-a|$$ As $N$ is arbitrary we get $\displaystyle \lim_{x\to 0^+}g(x)=0.$
When $a_n\to \infty$ we get $$xf(x)\ge x\sum_{n=N}^\infty e^{-nx}a_n\ge \left (\inf_{n\ge N}a_n\right )\ x\sum_{n=N}^\infty e^{-nx}\\ =\left (\inf_{n\ge N}a_n\right )\ {x\over 1-e^{-x}}e^{-Nx}$$ Taking the limit gives $$\liminf_{x\to 0^+}xf(x)\ge \inf_{n\ge N}a_n$$ As $N$ is arbitrary we get $\displaystyle \lim_{x\to 0^+} xf(x)=\infty.$
Remark The assumptions can be relaxed to the existence of the limit $$\lim_{n\to \infty}{a_0+a_1+\ldots +a_n\over n+1}$$