# Show that $\lim_{n \to \infty}\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} = \frac{a}{2}$

Assume $a_n \to a$. Then Show that $$\lim_{n \to \infty}\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} = \frac{a}{2}$$

So as $a_n \to a$ we know $\forall \epsilon, \exists N, \forall n \ge N ,|a_n - a | < \epsilon$.

Further, $\exists M > 0$ such that $|a_n| < M, \forall n$.

I want to do something similar to the proof for Cesaro means but it doesn't quite work out. We have:

$\displaystyle\lim_{n \to \infty }\bigg | \frac{na_1 + (n-1)a_2 + ... +2a_{n-1} + a_n}{n^2} - \frac{a}{2} \bigg| = \, \lim_{n \to \infty }\bigg | \frac{2na_1 + 2(n-1)a_2 + ... +4a_{n-1} + 2a_n - n^2a}{2n^2} \bigg| = \\ \displaystyle \frac{1}{2}\lim_{n \to \infty }\bigg | \frac{(2na_1 - na) + (2(n-1)a_2 - na) + ... + (4a_{n-1} - na) + (2a_n - na)}{n^2} \bigg| = \\ \displaystyle\frac{1}{2}\lim_{n \to \infty} \bigg | \frac{n(2a_1 - a) + n(2\frac{(n-1)}{n}a_2 - a) + ... + n(\frac{4}{n}a_{n-1} - a) + n(\frac{2}{n}a_n - a)}{n^2} \bigg| = \\ \displaystyle\frac{1}{2}\lim_{n \to \infty} \bigg | \frac{(2a_1 - a) + (2\frac{(n-1)}{n}a_2 - a) + ... + (\frac{4}{n}a_{n-1} - a) + (\frac{2}{n}a_n - a)}{n} \bigg|$.

I can continue but I am not sure if this is exactly the right way to go about this.

This is just a strict application of Cesaro's summation. See here General Cesaro $\lim\limits_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k}\sum_\limits{k=0}^{n}\lambda_k a_k =\lim\limits_{n\to\infty} a_n$ $$\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} =\sum_{i=1}^n \frac{(n+1)a_i}{n^2} -\sum_{i=1}^n \frac{ia_i}{n^2}\\\frac{n+1}{n} \left(\frac{1}{n}\sum_{i=1}^n a_i\right)- \frac{n+1}{2n}\left(\frac{1}{\sum_{i=1}^n i } \sum_{i=1}^n ia_i\right)\to a-\frac{a}{2}=\frac{a}{2}$$ since with $\lambda_i=i,1$ we have $$\sum_{i=1}^n i = \frac{n(n+1)}{2}~~~and ~~~\sum_{i=1}^n 1 = n$$

• I had never seen the general form of Cesaro. – oliverjones Dec 25 '17 at 2:15

$$\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} =\sum_{i=1}^n \frac{(n+1)a_i}{n^2} -\sum_{i=1}^n \frac{ia_i}{n^2}=\sum_{i=1}^n \frac{ia_i}{n^2}$$

$$\sum_{i=1}^n \frac{(n+1)a_i}{n^2} =\frac{(n+1)}{n^2}\sum_{i=1}^na_i=2\sum_{i=1}^n \frac{ia_i}{n^2}$$

$$\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} =\frac12 \frac{(n+1)}{n^2}\sum_{i=1}^na_i=\frac12 \frac{(n+1)}{n}\frac{\sum_{i=1}^na_i}{n}\to \frac{a}{2}$$

Indeed by Stolz–Cesàro theorem

$$\lim_{n\to+\infty} \frac{\sum_{i=1}^na_i}{n}=\lim_{n\to+\infty} \frac{\sum_{i=1}^{n+1}a_i-\sum_{i=1}^na_i}{n+1-n}=\lim_{n\to+\infty} a_{n+1}=a$$

• You were right (almost) except for the last step. Use the fact that $(1/n)\sum a_i\to a$ via Cesaro-Stolz. And then you can see that the expression is $(1/2)((n+1)/n)(1/n)\sum a_i\to (1/2)(1)(a)$. – Paramanand Singh Dec 25 '17 at 4:23
• @ParamanandSingh Thanks I've fixed this formal step. Why the asymptotic approch is considered not correct in this case? – user Dec 25 '17 at 8:15
• You just need to show that it is correct by mentioning the name Cesaro-Stolz. That's it +1 – Paramanand Singh Dec 25 '17 at 9:09
• @ParamanandSingh Thanks! fixed the $\sim$ – user Dec 25 '17 at 9:11

$\displaystyle \sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} = \sum_{i=1}^n \frac{a_i}{n} + \sum_{i=1}^n \frac{a_i}{n^2} - \sum_{i=1}^n \frac{ia_i}{n^2}$

Therefore $\displaystyle \lim_{n \to \infty}\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} = \lim_{n \to \infty} \sum_{i=1}^n \frac{a_i}{n} + \lim_{n \to \infty}\sum_{i=1}^n \frac{a_i}{n^2} - \lim_{n \to \infty}\sum_{i=1}^n \frac{ia_i}{n^2}$

The first limit is $a$ (arithmetic mean of a convergent sequence) the second one is $0$ and for the last one use Stolz–Cesàro theorem