Variables defined as floor and fraction part from exponentially distributed random variable

Since the pdf of $X$ is $\lambda e^{-\lambda x}\mathbb{1}_{x\geq 0}$ we have: $$\forall m\in\mathbb{N},\quad\mathbb{P}[\lfloor X\rfloor\leq m]=\int_{0}^{m+1}\lambda e^{-\lambda x}\,dx = 1-e^{-\lambda(m+1)}$$ hence the pdf of $\lfloor X \rfloor$ is $(e^{\lambda}-1)e^{-\lambda(m+1)}\mathbb{1}_{m\in\mathbb{N}}$ and: $$\mathbb{E}[\lfloor X\rfloor]=(e^{\lambda}-1)\sum_{m=0}^{+\infty} m e^{-\lambda(m+1)}=\frac{1}{e^\lambda-1}$$ since for any $-1<y<1$: $$\sum_{m=0}^{+\infty} m y^m = y\cdot\frac{d}{dy}\sum_{m=1}^{+\infty}y^m=\frac{y}{(1-y)^2}.$$ For the second part, we have: $$\forall t\in(0,1),\qquad \mathbb{P}[\{X\}\leq t]=\sum_{n\in\mathbb{N}}\int_{n}^{n+t}\lambda e^{-\lambda x}dx = (1-e^{-\lambda t})\sum_{n\in\mathbb{N}}e^{-\lambda n}=\frac{1-e^{-\lambda t}}{1-e^{-\lambda}}$$ hence the pdf of $\{X\}$ is $\frac{\lambda}{1-e^{-\lambda}}e^{-\lambda x}\mathbb{1}_{x\in(0,1)}$. By the linearity of $\mathbb{E}$, $$\mathbb{E}[\{X\}]=\mathbb{E}[X]-\mathbb{E}[\lfloor X\rfloor] = \frac{1}{\lambda}-\frac{1}{e^{\lambda}-1}.$$