Suppose $a_{n}>0$. $na_{n}$ is monotonic, and it approaches 0 as n approaches infinity. $\sum_{n=1}^{\infty} a_{n}$ is convergent. please prove $$ \lim_{n \to \infty} n\ln(n)\,a_{n}=0$$
I totally don't know how to solve it? Can somebody tell me how to solve it?
Maybe the problem is that how to handle the $\ln n$,if we can find a method cancel $\ln n$ ,then it will work.
On
Can you clarify whether this is a homework question?
I will reply as if it were. The following question and its answer will inspire you. Suppose that $a_n$ is a monotonous sequence, $a_n\geq 0$, such that $\sum a_n < \infty.$ Then
$$\lim na_n=0.$$
Indeed, the following tends to zero as $n\to \infty$:
$$a_{n+1}+\cdots +a_{2n},$$ and at the same time one sees that the above is
$$\geq n a_{2n},$$ and since $n a_{2n} \geq n a_{2n+1}$, we have proven $k a_k\to 0$ as $k$ goes to infinity.
This will help.
Suppose not. Then there is $\epsilon > 0$ such that $n \ln(n) a_n > \epsilon$ for infinitely many $n$. For any fixed positive integer $k$, take $n > k$ for which $n \ln(n) a_n > \epsilon$ and so large that $$\sum_{j=k}^n \dfrac{1}{j} \ge \dfrac{\ln(n)}2$$ For $k \le j \le n$ we have $a_j \ge n a_n/j$, so $$ \sum_{j=k}^n a_j \ge n a_n \sum_{j=k}^n \dfrac{1}{j} \ge \dfrac{n a_n \ln(n)}{2} > \dfrac{\epsilon}{2}$$ and therefore $\sum_j a_j$ diverges.