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

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

How should I start?

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

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

How should I start?

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

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

How should I start?

Maybe this might simplify something:

$\displaystyle \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) = \sum\limits_{n=1}^\infty \frac{1}{n} \left(\sum_{k = 1}^{2n-1} a_k - \sum_{k = 1}^{n-1} a_k \right)$

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

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**SlipEternal** Maybe this might simplify something:

$\displaystyle \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) = \sum\limits_{n=1}^\infty \frac{1}{n} \left(\sum_{k = 1}^{2n-1} a_k - \sum_{k = 1}^{n-1} a_k \right)$

So I understand how you get there I think. To parse this:

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

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

But I do not know where to go from there?

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

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**vidomagru** So I understand how you get there I think. To parse this:

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

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

But I do not know where to go from there?

Let $\displaystyle S_k = \sum_{n = 1}^k a_n$. Since the series converges absolutely, the sequence $\displaystyle S_k$ converges (and is Cauchy)[/tex]. Rewriting the series you are given:

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

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

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

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

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

Ok I follow your logic up until here:

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**SlipEternal** So, the general form of the series seems to be $\displaystyle \sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1}$. 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:

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

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

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 $\displaystyle \varepsilon>0$, there exists $\displaystyle N$ such that for all $\displaystyle n\ge N$, the partial sum of the first $\displaystyle n$ terms are no more than $\displaystyle \varepsilon$ apart?

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

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

I am not quite sure how to apply that, but I thought of something else instead that might be helpful. Isn't $\displaystyle (a_n + a_{n+1} + ... + a_{2n-1})$ a subsequence of $\displaystyle \sum\limits_{n=1}^\infty a_n$ which is convergent so any subsequence would also be convergent?

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

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

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.

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

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

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

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

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**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.

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

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

s

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

Also, writing out the terms of the series, we have:

$\displaystyle a_1 + \dfrac{a_2}{2} + \dfrac{5a_3}{6} + \dfrac{7a_4}{12} + \dfrac{47a_5}{60} + \ldots$

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

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

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**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.

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

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

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

I don't quite understand how $\displaystyle \underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}}$ is n terms? Also I cannot get induction to work for

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

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

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**SlipEternal** Also, writing out the terms of the series, we have:

$\displaystyle a_1 + \dfrac{a_2}{2} + \dfrac{5a_3}{6} + \dfrac{7a_4}{12} + \dfrac{47a_5}{60} + \ldots$

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

I also don't think I quite understand how you expanded the series.

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

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

$\displaystyle \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} +$$\displaystyle \ldots$

$\displaystyle a_1$ shows up in the first term only. $\displaystyle a_2$ shows up in the second term only. $\displaystyle a_3$ shows up in terms 2 and 3. $\displaystyle a_4$ shows up in terms 3 and 4. $\displaystyle a_5$ shows up in terms 3, 4, and 5. $\displaystyle a_6$ 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.

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

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

$\displaystyle \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} +$$\displaystyle \ldots$

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

Ok I see that and I understand that the coefficients follow $\displaystyle \dfrac{2n+1}{n(n+1)} \le 1$. Thus $\displaystyle \sum_{k = n}^{2n-1} a_k \le na_n, \forall n$. But where does this lead us now?