1. ## Series convergence

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

2. $\sum \frac{1}{n}$ is divergent, so for all $\varepsilon>0$, if $n$ is enough big,
$|a_n|\leq \frac{\varepsilon}{n}$, ie $|n a_n|\leq \varepsilon$.

3. I do not uderstand why we look at 1/n since it diverges,and in the problem series converges

4. Originally Posted by drpawel
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 $a_n = \begin{cases}1/\sqrt n&\text{if n=k^4 for some integer k},\\0&\text{otherwise},\end{cases}$
Then $\sum a_n = \sum_k 1/k^2$, which converges. But if $n=k^4$ then $na_n=\sqrt n$, so $a_n\not\to0$.

You need an extra condition such as requiring that the sequence (a_n) is decreasing, in order for the result to be true.

5. Thabk you for that correction.That sequence is indeed decreasing sequence of positive numbers.So,how to prove it.

6. Originally Posted by drpawel
Yes, indeed the sequence is decreasing sequence of the positive numbers.How can i prove that now?
I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.

Let $\varepsilon>0$. By the Cauchy criterion for convergence, there exists an integer N such that $\sum_{j=m}^na_j<\varepsilon/2$ whenever n>m>N. In particular, $\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$ for all p>0. If the sequence $(a_n)$ is decreasing, then $pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$, and therefore $(N+p)a_{N+p}<\frac{N+p}p\varepsilon/2$. But if p>N then $\frac{N+p}p<2$. It follows that if n>2N then $na_n<\varepsilon$. Therefore $na_n\to0$ as $n\to\infty$.

7. Originally Posted by drpawel
and therefore
If you have a specific follow-up question you will need to state it more clearly than this ....

8. Originally Posted by mr fantastic
If you have a specific follow-up question you will need to state it more clearly than this ....
I have some problems with my computer,I am trying to fix it, I did not do this by purpose.I appologise.

9. Originally Posted by Opalg
I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.

Let $\varepsilon>0$. By the Cauchy criterion for convergence, there exists an integer N such that $\sum_{j=m}^na_j<\varepsilon/2$ whenever n>m>N. In particular, $\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$ for all p>0. If the sequence $(a_n)$ is decreasing, then $pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$, and therefore $(N+p)a_{N+p}<\frac{N+p}p\varepsilon/2$. But if p>N then $\frac{N+p}p<2$. It follows that if n>2N then $na_n<\varepsilon$. Therefore $na_n\to0$ as $n\to\infty$.
Can you plese explain why (N+p)asub(N+p)<(N+p)/P, what about epsilon/2.Is it multiply time (N+p)/P.
Plese explain a little bit since I am lost.

10. Originally Posted by drpawel
Can you plese explain why (N+p)asub(N+p)<(N+p)/P, what about epsilon/2.Is it multiply time (N+p)/P.
Plese explain a little bit since I am lost.
In the sum $\sum_{j=N+1}^{N+p}a_j$, each term is greater than or equal to the last term (since the sequence $(a_j)$ is decreasing). So the sum is $\geqslant pa_{N+p}.$ That gives the inequality $pa_{N+p}<\varepsilon/2.$

But we're interested in what happens to $na_n$. So it would be more useful to know something about $(N+p)a_{N+p}$ rather than $pa_{N+p}$. That's the reason for multiplying both sides of the previous inequality by $\frac{N+p}p$, getting $(N+p)a_{N+p}<\frac{N+p}p\varepsilon/2.$

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.