uniform convergence $\sum_{n\ge 0} \ln(1+e^{-nx})$

Question

how prove uniform convergence : $$x\in (0,+\infty) \\ \sum_{n\ge 0} \ln(1+e^{-nx})$$

we have : $$\ln(1+e^{-nx}) \le e^{-nx}$$

and series :$\sum_{n\ge 0} e^{-nx} $ converge (série géométrique) so:series $\sum_{n\ge 0} \ln(1+e^{-nx})$ Uniform convergence

that is Corect or Not ?

if not correct . How prove Uniforme convergence .

Answer 1

No, your argument is not correct. You are right that $$ 0 \le \ln(1+e^{-nx}) \le e^{-nx} $$ for all $x$ and $n$, and that $\sum_n e^{-nx}$ converges for all $x$, but that does not prove the uniform convergence. You probably want to apply the Weierstrass M-test, but for that you need a sequence $M_n$ with $$ 0 \le \ln(1+e^{-nx}) \le M_n $$ and $\sum_n M_n < \infty$. The important point is that the bound $M_n$ does not depend on $x$.

Actually the series is not uniformly convergent. With $f_n(x) = \ln(1+e^{-nx})$ we have $$ f_n(1/n) = \ln(1+1/e) > 0 $$ so that $(f_n)$ does not uniformly converge to zero. Consequently, $\sum_n f_n(x)$ is not uniformly convergent.

Answer 2

For every $n \in \mathbb{N}$, let $\displaystyle R_n : x \longmapsto \sum_{k=n+1}^{+\infty} \ln(1+e^{-kx})$.

To see whether the series is uniformly convergent, we have to see if the sequence $(R_n)_{n \geq 0}$ converges uniformly to the null function.

But for every $n \in \mathbb{N}$, one has \begin{align*} ||R_n||_\infty \geq R_n\left( \dfrac{1}{n+1}\right) &= \sum_{k=n+1}^{+\infty} \ln(1+e^{-k/(n+1)}) \\ &= \ln(1+e^{-1})+\sum_{k=n+2}^{+\infty} \ln(1+e^{-k/(n+1)}) \\ &\geq \ln(1+e^{-1}) \end{align*}

Hence $||R_n||_\infty$ does not tend to $0$ as $n\rightarrow +\infty$, and the series is not uniformly convergent over $(0,+\infty)$.