# 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.

• I might be making a mistake somehow, but from your definitions, if we check the basic cases, it seems that: $S_0 - \sigma_0 = 0$ (as we want) but, $S_1 = a_0 + a_1$ and $\sigma_1 = S_0 + S_1 = 2a_0 + a_1$ (according to the definitions), thus $S_1-\sigma_1 = -a_0$ where the formulas rhs would be $\frac{a_1}{2}$. Which would indicate some kind of typo. – zo0x Apr 12 '17 at 14:00
• Yeah. I think there's some typo in this exercise. I wonder what the correct formula was supposed to be... – Dylan Apr 12 '17 at 14:04

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

$$\sigma_n =\sum_{k=0}^n\sum_{j=0}^na_k 1_{(k \leqslant j)} = \sum_{k=0}^n\sum_{j=k}^na_k = \sum_{k=0}^n a_k (n - k +1) = (n+1) \sum_{k=0}^n a_k - \sum_{k=0}^n k a_k \\\implies \sigma_n =(n+1)S_n - \sum_{k=0}^n k a_k \\ \implies S_n - \frac{\sigma_n}{n+1} = \frac{1}{n+1}\sum_{k=0}^n ka_k$$

• Thank you! I knew the author must have just left off some factor. +1 and accepted. – Dylan Apr 12 '17 at 16:22

For $n=0$,

$S_0=a_0$ and $\sigma_0=S_0=a_0$

thus

$$S_0-\sigma_0=0$$

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

$$S_n-\sigma_n=-\sigma_{n-1}.$$

• I saw that. See edits 1 and 2. Any idea what the correct identity should be? I suspect there's some small typo. – Dylan Apr 12 '17 at 14:10