Proving $S_n - \sigma_n = \frac{1}{n+1}\sum_{j=1}^nja_j$ Let $S_n = \sum_{j=0}^n a_j$ and $\sigma_n = \sum_{j=0}^n S_j$ for all $n$.
I'm trying to show that
$$S_n - \sigma_n = \frac{1}{n+1}\sum_{j=1}^nja_j$$
Here's how far I've been able to get:
$$\begin{align}S_n - \sigma_n &= \sum_{k=0}^na_k - \sum_{j=0}^n\sum_{k=0}^ja_k \\
&= \sum_{j=0}^n\frac{1}{n+1}\sum_{k=0}^na_k - \sum_{j=0}^n\sum_{k=0}^ja_k \\ 
&= \sum_{j=0}^n\frac{1}{n+1}\left[\sum_{k=0}^j\big(a_k - (n+1)a_k\big) + \sum_{k=j+1}^na_k\right] \\
&= \sum_{j=0}^n\frac{1}{n+1}\left[-n\sum_{k=0}^ja_k + \sum_{k=j+1}^na_k\right] \\
\end{align}$$
Comparing this to the result I'm supposed to get, I should have $$-n\sum_{k=0}^ja_k + \sum_{k=j+1}^na_k = ja_j$$ But I'm stuck on this last bit.  Any hints?

Edit: Actually, I just listed out the terms and I'm not even sure this formula is correct.  I'm getting
$$\begin{align}S_n - \sigma_n &= (a_0 + a_1 + a_2 + \cdots + a_n) - \big[(a_0) + (a_0 + a_1) + (a_0 + a_1 + a_2) + \cdots + S_n\big] \\ &= -na_0-(n-1)a_1 - \cdots -a_{n-1} \\ &= -\sum_{j=0}^n(n-j)a_j\end{align}$$
Can someone confirm if the above formula is correct or if the authors made some kind of typo in this question?

Edit 2: It's clear that there is a typo in this exercise.  Unless someone can discern what the identity is actually supposed to be, I'm going to delete this question in a few minutes and move on.
 A: For $n=0$,
$S_0=a_0$  and  $\sigma_0=S_0=a_0$
thus
$$S_0-\sigma_0=0$$
but your formula gives nothing.
For $n=1$,
$S_1=a_0+a_1$ and $\sigma_1=2a_0+a_1$
thus
$$S_1-\sigma_1=-a_0$$
and the formula gives $\frac {a_1}{2} $.
We should have
$$S_n-\sigma_n=-\sigma_{n-1}. $$
A: The "correct" formula may be a useful identity relating partial and Cesaro sums:
$$S_n - \frac{\sigma_n}{n+1} = \frac{1}{n+1}\sum_{j=0}^n ja_j$$
We prove this as follows.
$$\sigma_n = \sum_{j=0}^nS_j = \sum_{j=0}^n\sum_{k=0}^ja_k = \sum_{j=0}^n\sum_{k=0}^na_k 1_{(k \leqslant j)}= \sum_{k=0}^n\sum_{j=0}^na_k 1_{(k \leqslant j)}$$
where we use the indicator $1_{(k \leqslant j)} = 1 (\,\,\text{if} \,\,k \leqslant j), = 0 (\,\,\text{if} \,\,k > j)$ and switch the sums.
Continuing we get
$$\sigma_n =\sum_{k=0}^n\sum_{j=0}^na_k 1_{(k \leqslant j)} = \sum_{k=0}^n\sum_{j=k}^na_k  = \sum_{k=0}^n a_k (n - k +1) = (n+1) \sum_{k=0}^n a_k - \sum_{k=0}^n k a_k \\\implies \sigma_n =(n+1)S_n  - \sum_{k=0}^n k a_k \\ \implies S_n - \frac{\sigma_n}{n+1} = \frac{1}{n+1}\sum_{k=0}^n ka_k$$
