if a1+a2+a3.. decreasing sequence of positive numbers
How to prove that if a1+a2+a3 ... converges then lim (if n to infinity) of na(subn)=0
This result is not true without some further restriction on the terms a_n. For example, if $\displaystyle a_n = \begin{cases}1/\sqrt n&\text{if $n=k^4$ for some integer $k$},\\0&\text{otherwise},\end{cases}$
Then $\displaystyle \sum a_n = \sum_k 1/k^2$, which converges. But if $\displaystyle n=k^4$ then $\displaystyle na_n=\sqrt n$, so $\displaystyle a_n\not\to0$.
You need an extra condition such as requiring that the sequence (a_n) is decreasing, in order for the result to be true.
I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.
Let $\displaystyle \varepsilon>0$. By the Cauchy criterion for convergence, there exists an integer N such that $\displaystyle \sum_{j=m}^na_j<\varepsilon/2$ whenever n>m>N. In particular, $\displaystyle \sum_{j=N+1}^{N+p}a_j<\varepsilon/2$ for all p>0. If the sequence $\displaystyle (a_n)$ is decreasing, then $\displaystyle pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$, and therefore $\displaystyle (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2$. But if p>N then $\displaystyle \frac{N+p}p<2$. It follows that if n>2N then $\displaystyle na_n<\varepsilon$. Therefore $\displaystyle na_n\to0$ as $\displaystyle n\to\infty$.
In the sum $\displaystyle \sum_{j=N+1}^{N+p}a_j$, each term is greater than or equal to the last term (since the sequence $\displaystyle (a_j)$ is decreasing). So the sum is $\displaystyle \geqslant pa_{N+p}.$ That gives the inequality $\displaystyle pa_{N+p}<\varepsilon/2.$
But we're interested in what happens to $\displaystyle na_n$. So it would be more useful to know something about $\displaystyle (N+p)a_{N+p}$ rather than $\displaystyle pa_{N+p}$. That's the reason for multiplying both sides of the previous inequality by $\displaystyle \frac{N+p}p$, getting $\displaystyle (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2.$
I hope that makes it a bit clearer. As I said in my previous comment, this proof is fairly heavy-duty analysis. As far as I know, there are no easy shortcuts for proving this result.