Uniform convergence $\sum _{n=1}^{\infty }\:e^{-nx}$ on $D=[0.1,\infty)$

Question

Uniform convergence $\sum _{n=1}^{\infty }\:e^{-nx}$ on $D=[0.1,\infty)$

$$\sum _{n=1}^{\infty }\:e^{-nx}=\sum _{n=1}^{\infty }\:(e^{-x})^n$$

so its geometric series then $$\sum _{n=1}^{\infty }\:(e^{-x})^n=\frac{e^{-x}}{1-e^{-x}}$$ now we need to find $S(x)$ $$S(x)=\lim _{n\to \infty}S_n(x)=\lim _{n\to \infty}\frac{e^{-x}}{1-e^{-x}}$$

so how can i find this and check uniform convergence?

thanks a lot

Answer 1

One has, as $N \to \infty$, $$ \sup_{x \in D}\left|S(x)-S_N(x) \right|=\sup_{x \in D}\left|\frac{e^{-(N+1)x}}{1-e^{-x}}\right|\le \frac{e^{-(N+1)0.1}}{1-e^{-0.1}}\to 0 $$giving the desired conclusion.

Answer 2

Actually,$$S_n(x)=\sum_{k=1}^n(e^{-x})^k=\frac{e^{-x}-(e^{-x})^{n+1}}{1-e^{-x}}$$and, since $e^{-x}<1$ (for each $x\in[0.1,+\infty)$),$$\lim_{n\to\infty}S_n(x)=\frac{e^{-x}}{1-e^{-x}}.$$The convergence is uniform because, for each $x\in[0.1,+\infty)$, and each $n\in\mathbb N$, $(e^{-x})^n\leqslant(e^{-0.1})^n$ and the series $\sum_{n=1}^\infty(e^{-0.1})^n$ converges. So, all it takes is to apply the Weierstrass $M$-test.

Answer 3

We know that $f(t)=\sum_{n\geq 1}t^n$, a power series, converges uniformly on $[0,1-\epsilon]$ for all $\epsilon >0$, as its radius of converges is $1$.

So, the series $f(e^{-x})$ converges uniformly on $[\epsilon, \infty)$, choosing $\epsilon =0.1$ concludes the problem.