a sequence $\{s_n\}$ with $\sum s_n$ convergent what would be $\{s_n\ge0\}$  such that 
$$\sum_{n=1}^\infty s_n$$ converges
but 
$$\lim_{n\to\infty}(n s_n) \neq 0$$
 A: An counter-example inlovning positive and negative terms is when 
$a_n \frac{(-1)^n}{\sqrt{n}}$
the convegence can be check using the Alternating series test.
For an example with positive terms choose 
$a_n= \frac{1}{n}$ when $n$ is perfect square 
otherwise let it be $0$
A: Answers are up, this is more a meta-answer.
The problem is false as presented.  The notation "$\displaystyle \lim_{n \to \infty} na_n \neq 0$" means that the limit exists and is nonzero, and if (as stated in the question) the terms of the sequence are positive, that implies $\sum a_n$ is infinite.   The correct statement is "$na_n$ does not converge to zero", which includes the case where no limit exists.  
The similarity between the posted answers is explained by the fact that in any solution there exists an infinite subsequence where $na_n$ is at least equal to a constant but for which $\sum a_n$ restricted to the subsequence is finite.  For construction of examples, it is necessary and sufficient to specify such a subseries, and fill the rest of the terms with any convergent series.
A: $a_n=\frac{1}{n}$ if $n$ is a perfect square and $a_n=\frac{1}{n^2}$ otherwise. 
P.S. This example is generic in the following sense:
Lemma If $a_n \geq 0, \sum a_n$ is convergent and $\lim_n na_n$ exists then $\lim na_n=0$.
Proof: Assume by contradiction that $\lim_n na_n \neq 0$. Then there exists some $M>0$ so that $na_n >M$ for all $n>N$. Then 
$$a_n >M \frac{1}{n} \forall  n >N \Rightarrow \sum_a_n =\infty$$
since the harmonic series diverges.

Thus, the only way you can construct an example as you want is by trying to make $\lim_n na_n$ not to exists.
A: Let $a_n=(-1)^{n+1} \frac {1}{n},\, n=1, 2, 3, \ldots.$
Then $ \sum_{n=1}^{\infty} a_n =\ln 2$.
