# Convergence of a sequence of partial sums of partial sums [duplicate]

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.

• 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. Nov 6, 2018 at 22:47
• It is the same question with different glasses. There $a_n$, $s_n$, here $s_n$, $\sigma_n$. Nov 6, 2018 at 22:50
• 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?) Nov 6, 2018 at 22:51

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 $$L=\lim a_n=\sum x_n$$.

• 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 $\frac{1}{n} s_n$. Nov 6, 2018 at 23:04

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

then

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