Show that the following series converges uniformly on $D$ = {$x\in \mathbb{R}:ln(0.6)\leq x$}: $\sum_{n=1}^{\infty}\frac{e^{-nx}}{2^{n}}$
My approach:
For $x\in D$, the maximum value of $f_n(x)=\frac{e^{-nx}}{2^{n}}$ is when $x$ is small, and the smallest $x$ on $D$ is $ln(0.6)$. So, the max value of $f_n(x)$ is:
$f_n(ln(0.6)) =\frac{e^{-nln(0.6)}}{2^n}=\frac{1}{e^{nln(0.6)}2^n} = \frac{1}{(e^{ln(0.6)})^{n}2^n} = \frac{1}{(1.2)^n}$.
So,
$|f_n(x)| \leq |\frac{1}{(1.2)^n}|$ for all $x\in D$.
Now, if we let $M=\frac{1}{(1.2)^n}$, then $|f_n(x)|\leq M_n$ for all $x\in D$.
$\sum_{n=1}^{\infty}\frac{1}{(1.2)^n}$ converges absolutely, so by M-test, $\sum_{n=1}^{\infty}\frac{e^{-nx}}{2^{n}}$ converges uniformly on $D$.
Is this right?