I wish to show that the series $\displaystyle \sum_{k=0}^{\infty} \frac{\sqrt{1-e^{-k}}}{3^k}$ converges.
Direct comparison test: if $\sum b_n$ converges and $0 \le a_n \le b_n$ for all $n$ then $\sum a_n$ also converges.
Since $e^{-k} > 0$ for all $k$, we have $-e^{-k} < 0$ hence $1-e^{-k} < 1$ and thus $\sqrt{1-e^{-k}} < 1$for all $k \in \mathbb{N}$.
Hence $ \displaystyle \frac{\sqrt{1-e^{-k}}}{3^k} \le \frac{1}{3^k}$ and $\displaystyle \sum_{k=0}^{\infty} \frac{1}{3^k}$ converges (being a geometric series). Therefore our original series converges.
Could someone please confirm whether or not this one is correct?
Your comparison procedure is fine. If you like to achieve extra accuracy in approximating the given series, you may notice that for any $x\in(0,1)$ we have $\sqrt{1-x}\leq 1-\frac{x}{2}$ and such inequality is pretty tight if $x$ is close to the origin. In particular $$ \sum_{k\geq 0}\frac{\sqrt{1-e^{-k}}}{3^k}=\sum_{k\geq 1}\frac{\sqrt{1-e^{-k}}}{3^k}\leq\sum_{k\geq 1}\frac{1}{3^k}-\frac{1}{2}\sum_{k\geq 1}\frac{1}{(3e)^k}=\frac{1}{2}-\frac{1}{2(3e-1)}\leq\frac{4}{9} $$ where the RHS is a pretty accurate approximation of the LHS.