I have to find a sequence ${\{f_k\}}_{k = 1}^{\infty}$, with $f_k : \mathbb{N} \to [0 , \infty)$, such that $\lim_{k \to \infty} f_k(n) = 0$ for all $n \in \mathbb{N}$ and $\int_{\mathbb{N}} f_k d \mu = 1$ for all $k \in \mathbb{N}$, where $\mu$ is the counting measure, i.e. $\sum_{n = 1}^{\infty} f_k(n) = 1$ for all $k \in \mathbb{N}$.
My main attempt was to write $f_k(n) = \frac 1 {k 2^n}$, what gives me $\lim_{k \to \infty} f_k(n) = 0$ for all $n \in \mathbb{N}$, but $\sum_{n = 1}^{\infty} f_k(n) = \frac 1 k \neq 1$ if $k \neq 1$, then it is not valid. The two conditions seem to be incompatible.
One solution is $$f_k(n)=(e^k-1)e^{-kn}$$ Then $$\lim_{k\rightarrow+\infty}f_k(n)=0$$ and $$\sum_{n=1}^{+\infty}f_k(n)=(e^k-1)\sum_{n=1}^{+\infty}e^{-kn}=\frac{(e^k-1)e^{-k}}{1-e^{-k}}=1$$