Math Help - Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

1. Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Let $\sum\limits_{n=1}^\infty a_n$ be a convergent series in which $a_n \ge 0, \forall n=1,2, ...$. Prove that the series $\sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1})$ converges as well.

How should I start?

2. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by vidomagru
Let $\sum\limits_{n=1}^\infty a_n$ be a convergent series in which $a_n \ge 0, \forall n=1,2, ...$. Prove that the series $\sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1})$ converges as well.

How should I start?

3. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by vidomagru
Let $\sum\limits_{n=1}^\infty a_n$ be a convergent series in which $a_n \ge 0, \forall n=1,2, ...$. Prove that the series $\sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1})$ converges as well.

How should I start?
Maybe this might simplify something:
$\sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) = \sum\limits_{n=1}^\infty \frac{1}{n} \left(\sum_{k = 1}^{2n-1} a_k - \sum_{k = 1}^{n-1} a_k \right)$

4. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
Maybe this might simplify something:
$\sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) = \sum\limits_{n=1}^\infty \frac{1}{n} \left(\sum_{k = 1}^{2n-1} a_k - \sum_{k = 1}^{n-1} a_k \right)$
So I understand how you get there I think. To parse this:

I know that $\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges.

I believe that $(\sum_{k = 1}^{2n-1} a_k$ and $\sum_{k = 1}^{n-1} a_k)$ converge since $\sum\limits_{n=1}^\infty a_n$ in convergent for all $n = 1,2,...$. Is that right?

But I do not know where to go from there?

5. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by vidomagru
So I understand how you get there I think. To parse this:

I know that $\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges.

I believe that $(\sum_{k = 1}^{2n-1} a_k$ and $\sum_{k = 1}^{n-1} a_k)$ converge since $\sum\limits_{n=1}^\infty a_n$ in convergent for all $n = 1,2,...$. Is that right?

But I do not know where to go from there?
Let $S_k = \sum_{n = 1}^k a_n$. Since the series converges absolutely, the sequence $S_k$ converges (and is Cauchy)[/tex]. Rewriting the series you are given:

$\sum_{n = 1}^\infty \dfrac{S_{2n-1} - S_{n-1}}{n}$

Writing out a few elements of this series, you have $S_1 + \dfrac{S_3 - S_1}{2} + \dfrac{S_5 - S_2}{3} + \dfrac{S_7 - S_3}{4} + \ldots$

Collecting like terms, we have $\dfrac{S_1}{2} - \dfrac{S_2}{3} + \dfrac{S_3}{4} - \dfrac{S_4}{5} \pm \ldots$

So, the general form of the series seems to be $\sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1}$. Use induction to prove that this is true, and then use the Alternating Series Test to prove that it converges conditionally.

6. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
So, the general form of the series seems to be $\sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1}$. Use induction to prove that this is true, and then use the Alternating Series Test to prove that it converges conditionally.
If I try to prove this is true using induction I have:

$\sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1}$ evaluated at n=1 gives $(-1)^{1+1} \dfrac{S_1}{1+1} = \dfrac{a_1}{2}$.

However this is not equal to $\sum\limits_{n=1}^\infty \dfrac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1})$ evaluated at $n=1$, i.e., $(a_1 + a_2 + ... + a_1)$.

7. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Hmm, that's a good point. Maybe you can show that for any $\varepsilon>0$, there exists $N$ such that for all $n\ge N$, the partial sum of the first $n$ terms are no more than $\varepsilon$ apart?

8. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
Hmm, that's a good point. Maybe you can show that for any $\varepsilon>0$, there exists $N$ such that for all $n\ge N$, the partial sum of the first $n$ terms are no more than $\varepsilon$ apart?
I am not quite sure how to apply that, but I thought of something else instead that might be helpful. Isn't $(a_n + a_{n+1} + ... + a_{2n-1})$ a subsequence of $\sum\limits_{n=1}^\infty a_n$ which is convergent so any subsequence would also be convergent?

Or because $(a_n + a_{n+1} + ... + a_{2n-1})$ is a partial sum of a convergent series $\sum\limits_{n=1}^\infty a_n$, $(a_n + a_{n+1} + ... + a_{2n-1})$ also converges.

Then we would just need to apply $\sum\limits_{n=1}^\infty \dfrac{1}{n}$. Not sure how to do that?

9. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

$\sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)$

Now, can you show that $\sum_{k = n}^{2n-1} a_k \le na_n$ for almost all $n$?

Since $\sum_{n\ge 1} a_n$ converges and $0\le a_n$ for all n, it must be that $\lim_{n\to \infty} a_n = 0$. So what I am getting at, $\underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}} \le \underbrace{a_n + a_n + \ldots + a_n}_{n\mbox{ terms}}$

10. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

$\sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)$

Now, can you show that $\sum_{k = n}^{2n-1} a_k \le na_n$ for almost all $n$?
s

11. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Also, writing out the terms of the series, we have:
$a_1 + \dfrac{a_2}{2} + \dfrac{5a_3}{6} + \dfrac{7a_4}{12} + \dfrac{47a_5}{60} + \ldots$

Can you show that the coefficient of $a_n$ is always no greater than 1?

12. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

$\sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)$

Now, can you show that $\sum_{k = n}^{2n-1} a_k \le na_n$ for almost all $n$?

Since $\sum_{n\ge 1} a_n$ converges and $0\le a_n$ for all n, it must be that $\lim_{n\to \infty} a_n = 0$. So what I am getting at, $\underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}} \le \underbrace{a_n + a_n + \ldots + a_n}_{n\mbox{ terms}}$
I don't quite understand how $\underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}}$ is n terms? Also I cannot get induction to work for

Originally Posted by SlipEternal
Now, can you show that $\sum_{k = n}^{2n-1} a_k \le na_n$ for almost all $n$?

13. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
Also, writing out the terms of the series, we have:
$a_1 + \dfrac{a_2}{2} + \dfrac{5a_3}{6} + \dfrac{7a_4}{12} + \dfrac{47a_5}{60} + \ldots$

Can you show that the coefficient of $a_n$ is always no greater than 1?
I also don't think I quite understand how you expanded the series.

14. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

$\sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k = n}^{2n-1}a_k\right)$

$\dfrac{a_1}{1} + \dfrac{a_2 + a_3}{2} + \dfrac{a_3 + a_4 + a_5}{3} + \dfrac{a_4 + a_5 + a_6 + a_7}{4} + \dfrac{a_5 + a_6 + a_7 + a_8 + a_9}{5} +$ $\ldots$

$a_1$ shows up in the first term only. $a_2$ shows up in the second term only. $a_3$ shows up in terms 2 and 3. $a_4$ shows up in terms 3 and 4. $a_5$ shows up in terms 3, 4, and 5. $a_6$ shows up in terms 4, 5, and 6. Seeing a pattern?

Also, this expansion should show you how each term is a sum of n elements over n.

15. Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

Originally Posted by SlipEternal
$\sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k = n}^{2n-1}a_k\right)$

$\dfrac{a_1}{1} + \dfrac{a_2 + a_3}{2} + \dfrac{a_3 + a_4 + a_5}{3} + \dfrac{a_4 + a_5 + a_6 + a_7}{4} + \dfrac{a_5 + a_6 + a_7 + a_8 + a_9}{5} +$ $\ldots$

$a_1$ shows up in the first term only. $a_2$ shows up in the second term only. $a_3$ shows up in terms 2 and 3. $a_4$ shows up in terms 3 and 4. $a_5$ shows up in terms 3, 4, and 5. $a_6$ shows up in terms 4, 5, and 6. Seeing a pattern?
Ok I see that and I understand that the coefficients follow $\dfrac{2n+1}{n(n+1)} \le 1$. Thus $\sum_{k = n}^{2n-1} a_k \le na_n, \forall n$. But where does this lead us now?

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