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Let $(x_n)$ be a real-valued sequence with partial sums $s_n = x_1 + x_2 + ... +x_n$. We define $\sigma_n = \frac{1}{n}(s_1 + s_2 + ... + s_n)$. Now, supposing $\Sigma\,x_n$ is convergent, I need to show that the sequence $(\sigma_n)$ is convergent and that $lim\,\sigma_n = \Sigma\,x_n$.

My intuition is to simply say that since $\Sigma\,x_n$ converges, then $lim\,sup\,s_n = lim\,s_n = L$, some real number, and then try to find a way to use a convergence test.

But I don't have any information about the $x$-values, so I can't say they're non-negative, which means I can't use comparison, and the ratio and root test both come out inconclusive since $lim\,\frac{\sigma_{n+1}}{\sigma_n}=1$.

Should I be finding a way to use the Cauchy criterion here? Or am I missing a different technique altogether?

EDIT: This isn't actually a duplicate of the arithmetic mean problem, because there is only one series of partial sums in the arithmetic mean. This problem takes the sum of each of those sums from 1 to n and presents a new problem with that.

EDIT 2: Upon further review, although not identical, the problem turns out to essentially be a re-skin of the arithmetic mean. See answer below.

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  • $\begingroup$ This isn't quite a duplicate of the above problem, because there is only one series of partial sums in the arithmetic mean. This problem takes the sum of each of those sums from 1 to n and presents a new problem with that. $\endgroup$
    – notadoctor
    Nov 6, 2018 at 22:47
  • $\begingroup$ It is the same question with different glasses. There $a_n$, $s_n$, here $s_n$, $\sigma_n$. $\endgroup$ Nov 6, 2018 at 22:50
  • $\begingroup$ Here we have $x_n$, $s_n$, and $\sigma_n$ (that is, I don't know that $s_n$ converges, I only know that $x_n$ does... although maybe I can just use the technique from the arithmetic mean problem twice?) $\endgroup$
    – notadoctor
    Nov 6, 2018 at 22:51

2 Answers 2

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Following Convergence of the arithmetic mean

Let $a_n:=s_n=x_1+\cdots+x_n$.

$(a_n)_n$ is convergent, because $\sum x_n$ is convergent.

Hence, (by the previous post) $\frac{1}{n}(a_1+\cdots+a_n)$ is convergent to $L:=\lim a_n$.

But $\frac{1}{n}(a_1+\cdots+a_n)=\sigma_n$ , so $(\sigma_n)_n$ is convergent to $L=\lim a_n=\sum x_n$.

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    $\begingroup$ Thank you for clearing this up! I was stuck on not knowing whether $s_n$ was convergent or not, and it somehow didn't occur to me to simply ignore it and work directly with $\frac{1}{n} s_n$. $\endgroup$
    – notadoctor
    Nov 6, 2018 at 23:04
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By Stolz–Cesàro theorem

$$\lim_{n\to \infty }\frac{(s_1 + s_2 + ... + s_n+s_{n+1})-(s_1 + s_2 + ... + s_n)}{n+1-n}=\lim_{n\to \infty } s_{n+1}=\sum x_n$$

then

$$\lim_{n\to \infty }\sigma(n)=\lim_{n\to \infty }\frac{s_1 + s_2 + ... + s_n}{n}=\sum x_n$$

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