Let $(R_N)_N$ be the sequence of remainders of your series, namely
$$\forall N\in\mathbb{N},\ R_N=\sum_{n=N+1}^{+\infty}a_n.$$
Since your series converges, the sequence $(R_N)_N$ is well defined and
$$\lim_{N\to+\infty}R_N=0.$$
Now, since your $a_n$'s are non-negative and the sequence is non-increasing,
$$\forall n\in\mathbb{N},\ na_{2n}\leq a_{n+1}+\cdots+a_{2n}\leq R_n.$$
By the Squeeze Theorem,
$$\lim_{n\to+\infty}na_{2n}=0.$$
For the odd subsequence: write, for $n\in\mathbb{N}$,
$$(2n+1)a_{2n+1}=2na_{2n+1}+a_{2n+1}\leq2na_{2n}+a_{2n+1}$$
and conclude (by the Squeeze Theorem again) that
$$\lim_{n\to+\infty}(2n+1)a_{2n+1}=0.$$
Finally, you have a sequence $(na_n)_{n\in\mathbb{N}}$ such that the odd and even subsequences have a nil limit: you can conclude that the sequence $(na_n)_{n\in\mathbb{N}}$ has a nil limit.
Regarding your questions:
- Suppose I have $2na_{2n}\to0$ and $(2n+1)a_{2n+1}\to0$; is that enough to say that $na_n\to0$?
Yes: a sequence has a limit $\ell$ if and only if its odd and even subsequences have the same limit, equal to $\ell$. Apply this result to the sequence $(na_n)_n$.
- Is there any easy way to show $na_{2n}\leq a_{n+1}+\cdots+a_{2n}$, there's probably a simple inductive proof I couldn't get.
You don't need induction here. Since the sequence $(a_n)_n$ is non-decreasing,
$$\forall n\in\mathbb{N}^*,\ a_{n+1}\geq a_{n+2}\geq\cdots\geq a_{2n},$$
hence
$$a_{n+1}+\cdots+a_{2n}=\sum_{k=n+1}^{2n}a_k\geq\sum_{k=n+1}^{2n}a_{2n}=na_{2n}.$$
