Prove that if $\sum a_n$ converges, then $na_n \to 0$.

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  • If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$ [duplicate]

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Solutions Collecting From Web of "Prove that if $\sum a_n$ converges, then $na_n \to 0$."

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
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,

For the odd subsequence: write, for $n\in\mathbb{N}$,
and conclude (by the Squeeze Theorem again) that

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,
$$\forall n\in\mathbb{N}^*,\ a_{n+1}\geq a_{n+2}\geq\cdots\geq a_{2n},$$