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.

  • $\begingroup$ 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. $\endgroup$ – zo0x Apr 12 '17 at 14:00
  • $\begingroup$ Yeah. I think there's some typo in this exercise. I wonder what the correct formula was supposed to be... $\endgroup$ – 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

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

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

For $n=0$,

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



but your formula gives nothing.

For $n=1$,

$S_1=a_0+a_1$ and $\sigma_1=2a_0+a_1$



and the formula gives $\frac {a_1}{2} $.

We should have

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.