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

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

Originally Posted by vidomagru
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?
The coefficients follow $\dfrac{2n+1}{n(n+1)}$ when $n$ is even. But not when it is odd. Anyway, if you can show that the coefficients are no greater than 1, then you can show that $0\le \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k=n}^{2n-1} a_k \right) = \sum_{n\ge 1} b_na_n$ where $b_n = \begin{cases}\dfrac{2n+1}{n(n+1)} & n\mbox{ is even} \\ ??? & n\mbox{ is odd}\end{cases} \le 1$. Hence, $\sum_{n\ge 1} b_na_n \le \sum_{n\ge 1} a_n$

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

Originally Posted by SlipEternal
The coefficients follow $\dfrac{2n+1}{n(n+1)}$ when $n$ is even. But not when it is odd. Anyway, if you can show that the coefficients are no greater than 1, then you can show that $0\le \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k=n}^{2n-1} a_k \right) = \sum_{n\ge 1} b_na_n$ where $b_n = \begin{cases}\dfrac{2n+1}{n(n+1)} & n\mbox{ is even} \\ ??? & n\mbox{ is odd}\end{cases} \le 1$. Hence, $\sum_{n\ge 1} b_na_n \le \sum_{n\ge 1} a_n$
I am not sure that I got that right. For example when n = 2, $\dfrac{2(2) + 1}{2(2+1)} = \dfrac{5}{6}$ which is the coefficient for $a_3$.

Also a_1 and a_2 appear once, a_3 and a_4 appear twice, a_5 and a_6 appear three times, a_7 and a_8 appear four times, etc. I'm not quite sure how to fit that into the coefficients since the number of a_n terms for any given n increases every two terms.

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

$\begin{array}{lll}b_1 = 1, & b_2 = \dfrac{1}{2}, & b_3 = \dfrac{1}{2} + \dfrac{1}{3}, \\ \\ b_4 = \dfrac{1}{3} + \dfrac{1}{4}, & b_5 = \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}, & b_6 = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} \\ \\ b_7 = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7}, & b_8 = \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}, & b_9 = \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} \end{array}$

So, let's try to find a recurrence relation for $b_n$:

$b_2 = b_1 - \dfrac{1}{1} + \dfrac{1}{2} = b_1 - \dfrac{1}{2}$

$b_3 = b_2 + \dfrac{1}{3}$

$b_4 = b_3 - \dfrac{1}{2} + \dfrac{1}{4} = b_3 - \dfrac{1}{4}$

$b_5 = b_4 + \dfrac{1}{5}$

$b_6 = b_5 - \dfrac{1}{3} + \dfrac{1}{6} = b_5 - \dfrac{1}{6}$

In general, this recursion seems to be $b_{n+1} = b_n + \dfrac{(-1)^n}{n+1}$ with initial condition $b_1 = 1$.

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

So, if I am correct about $b_n$, then this is true:

$\sum_{n\ge 1} \dfrac{1}{n}(a_n + \ldots + a_{2n-1}) = \sum_{n\ge 1} \left(a_n\sum_{k = 1}^n \dfrac{(-1)^{k+1}}{k}\right)$

The sum $\sum_{k = 1}^n \dfrac{(-1)^{k+1}}{k}$ is an alternating harmonic series, so it reaches a maximum value of 1 and a minimum value of $\dfrac{1}{2}$. Its limit as $n\to \infty$ is $\ln 2$.

So, again, if $b_n$ is really a partial sum of the alternating harmonic series, then

$\dfrac{1}{2}\sum_{n\ge 1} a_n \le \sum_{n\ge 1} \dfrac{1}{n}(a_n + \ldots + a_{2n-1}) = \sum_{n \ge 1} \left(a_n\sum_{k = 1}^n \dfrac{(-1)^{k+1}}{k}\right) \le \sum_{n\ge 1} a_n$

So, it converges somewhere between 1/2 and 1 times the original series.

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