Find a function $f$ such that $f \in L_p$ if and only if $1 ≤ p ≤ p_0$ with certain measure. I am working on this question which is from Bartle's "The Elements of Integration and Lebesgue Measure" (questions 6.H and 6.I for reference).
Let $X = \mathbb{N}$ and let $\mathbb{X}$ be the collection of all subsets of N. Let $λ$ be defined by:
\begin{align*}
λ(E) = \sum_{n=1}^{\infty} \frac{1}{n^2}, E \in \mathbb{X}, n \in E
\end{align*}
$(a)$ Show that $f : X → R, f(n) = \sqrt{n}$ satisfies $f \in L_p$ if and only if $1 ≤ p < 2$.
$(b)$ Find a function $f$ such that $f \in L_p$ if and only if $1 ≤ p ≤ p_0$.
Part $(a)$ is simple enough by the divergence of harmonic series as;
\begin{align*}
\int |f|^p d\lambda = \sum_{n=1}^{\infty} \frac{|f(n)|^p}{n^2}
\end{align*}
From that it is also easy to show that, again by divergence of harmonic series, that for $f(n)=n^\frac{1}{p_o}$ if and only if $1 ≤ p < p_0$. 
I am stuck on trying to extend this to include when $p=p_0$. My first thought is to somehow change the function $f(n)=n^\frac{1}{p_o}$ such that series is the alternating harmonic series when $p = p_0$ and this still convergent, however this is tricky due to power involved. Any tips would be much appreciated. 
 A: Here's another idea: Let $(p_k)_{n=1}^{\infty}$ be a sequence in $(p_0,\infty)$ such that $p_k \to p_0$. Define functions $f_k(n) = n^{1/p_k}$ and let
$$f(n) = \sum_{k=1}^{\infty}\frac{1}{2^k\Vert f_k \Vert_{p_0}^{p_0}}f_k(n).$$
By what you have shown, $f_k \in L^p$ if and only if $1 \leq p < p_k$. In particular $f_k \in L^{p_0}$. Since $f_k(n) \geq 1$, for $1 \leq p \leq p_0$ we have $$\Vert f_k \Vert_p^p \leq \Vert f_k \Vert_{p_0}^{p_0}.$$
Taking $p$-th roots and noting that $\Vert f_k \Vert_{p_0} > 1,$ we see that
$$\Vert f_k \Vert_p \leq \Vert f_k \Vert_{p_0}^{p_0/p} \leq \Vert f_k \Vert_{p_0}^{p_0},$$
thus
$$\Vert f \Vert_p \leq \sum_{k=1}^{\infty} \frac{1}{2^k \Vert f_k \Vert_{p_0}^{p_0}} \Vert f_k \Vert_p \leq \sum_{k=1}^{\infty} \frac{1}{2^k} < \infty,$$
so $f \in L^p$ for $1 \leq p \leq p_0$. On the other hand, if $p > p_0$, then choosing $k$ such that $p_k < p$ (here we use $p_k \to p_0$), we have that $f_k \not\in L^p$. Then $f \not\in L^p$ since $f \geq \frac{1}{2^k \Vert f_k \Vert_{p_0}^{p_0}} f_k.$
