Prove $\lim\limits_{n\to\infty}na_n\ln(n)=0$

Lemma: If $\sum b_n $ converges and $b_n>0$ is monotone, then $n b_n \to 0 .$

Proof: By the Cauchy Condensation test we have $\sum 2^n b_{2^n} $ converges as well, so $2^n b_{2^n}\to 0.$ For $2^n < k < 2^{n+1} $ we have $k b_k \leq 2^{n+1} b_{2^n} \to 0.$ Thus $n b_n \to 0.$

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By the Cauchy Condensation test $\sum 2^n a_{2^n} $ converges, the lemma shows $ 2^n a_{2^n} \to 0$ and we are given that this occurs monotonically.

Thus applying our lemma to $\sum 2^n a_{2^n} $ gives us that $ n 2^n a_{2^n} \to 0.$ Now for $2^n < k < 2^{n+1} $ we have $$ \log_2 k\cdot k a_k \leq (n+1) 2^{n+1} a_{2^n} \to 0.$$

Thus $$ \lim_{n\to\infty} n a_n \log n = 0.$$

Since $(na_n)$ is positive and monotone, $(na_n)$ is nonincreasing and converges to zero hence $na_n=\sum\limits_{k\geqslant n}c_k$ for a nonnegative summable sequence $(c_k)$.

Since $\sum\limits_na_n$ converges, $\sum\limits_kc_kx_k$ converges, with $x_k=\log k$. Furthermore, $na_n\log n=\sum\limits_{k}c_kx_k(n)$ with $x_k(n)=[n\leqslant k]\log n$. Hence $x_k(n)\leqslant x_k$ and $x_k(n)\to0$ when $n\to\infty$. By dominated convergence, $na_n\log n\to0$.