If $s_n=\sum_{k=0}^{n}a_k$ and $n\log n \ a_n\rightarrow 0 \ (n\to \infty)$ show $\frac{s_n}{\log n}\to 0 \ (n\to \infty).$ Let $s_n=\sum_{k=0}^{n}a_k$ and $n\log n \ a_n\rightarrow 0 \ (n\to \infty).$ Is it true that $\frac{s_n}{\log n}\to 0 \ (n\to \infty)?$
 A: We see that the condition on the coefficients gives
$$\lim_{n\to\infty}{|a_n|\over 1/(n\log n)}=0$$
because if the limit is $0$, the absolute value of the limit is also $0$.
so that for $N>\!>0$ sufficiently large, we have $|a_n|<{1\over n\log n}$, by definition of a limit of a sequence as $n\to\infty$.
But then
$$\left|\sum_{k=N}^n a_n\right|<\sum_{k=N}^n{1\over k\log k}\le 2\log\log n$$
by using the integral test for sums of positive, monotone functions you learn in calculus.
Then we have

$${1\over \log n}|s_n|\le {C+2\log\log n\over\log n}\stackrel{n\to\infty}{\longrightarrow} 0.$$

with $C=\left|\sum_{k=0}^{N-1}a_k\right|$ to account for the first bunch of terms before we could guarantee $|a_n|<{1\over n\log n}$.
A: With Stolz-Cesaro's lemma, if $\lim \frac {s_n - s_{n-1}} {\log n - \log (n-1)} = l$ then $\lim \frac {s_n} {\log n} =l$. But the former limit is $\lim \frac {a_n} {\log \frac {n} {n-1}} = \lim \frac {a_n} {\frac 1 n} = \lim n a_n$ and $\lim |n a_n| \leq \lim |n a_n| \log n =0$ by hypothesis, so $\lim n a_n$ must be $0$ too. (I have also used that $\lim_{x \to 0} \frac {\log (1+x)} x = 1$, with $x = \frac 1 {n-1}$ and $\lim \frac {n-1} n =1$.)
