If $a_n\to a$ then $b_n=\frac{1}{n^2}\sum_{i=1}^n ia_i \to a/2$.

Suppose the sequence $\{a_n\}$ of real numbers has limit $a$. Define $b_n=\frac{1}{n^2}\sum_{i=1}^n ia_i$ and prove that $b_n\to a/2$.

My attempts: I first tried writing down the definition of the limit.

$$ \left|\frac{1}{n^2}\sum_{i=1}^n ia_i - a/2\right| $$

I don't think I have any great ideas about what to do here. I could distribute the $1/n^2$ but that doesn't seem useful. I could add and subtract terms looking for a way to apply the triangle inequality. But what terms? I thought about maybe $\sum_{i=1}^n a_i$ or something like that, but it doesn't seem useful as far as I can tell.

I also thought maybe the AM-GM inequality might be useful. If I were merely trying to prove convergence and not to some point that might be true, but I don't see how to use it here.

I thought this also kinda looks like something I've seen in Calc 2 where you take a derivative to find a Taylor expansion. But here there is no polynomial just an arbitrary sequence $\{a_n\}$.

So I'm out of ideas at this point.

Solution 1:

Let $\epsilon>0$ be given. By definition, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|a_n-a|<\epsilon$. This then implies $a-\epsilon<a<a+\epsilon$ and we have

$$\sum_{i=1}^n i a_i=\sum_{i=1}^N ia_i+\sum_{i=N+1}^n ia_i$$

$$\Rightarrow \sum_{i=N+1}^n i(a-\epsilon)<\sum_{i=1}^n i a_i-\sum_{i=1}^N ia_i<\sum_{i=N+1}^n i(a+\epsilon)$$

$$\Rightarrow (a-\epsilon)\frac{1}{2}(n-N)(n+N+1)<\sum_{i=1}^n i a_i-\sum_{i=1}^N ia_i<(a+\epsilon)\frac{1}{2}(n-N)(n+N+1)$$

This then implies

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

Taking limits as $n$ approaches infinity gives

$$\frac{a-\epsilon}{2}=\lim_{n\to\infty}(a-\epsilon)\frac{(n-N)(n+N+1)}{2n^2}$$

$$<\lim_{n\to\infty}b_n+\left[\frac{1}{n^2}\sum_{i=1}^N \right]ia_i\lim_{n\to\infty}\frac{1}{n^2}$$

$$<\lim_{n\to\infty}(a+\epsilon)\frac{(n-N)(n+N+1)}{2n^2}=\frac{a+\epsilon}{2}$$

Since the limit of $\frac{1}{n^2}$ is $0$ this simplifies to

$$\frac{a-\epsilon}{2}<\lim_{n\to\infty}b_n<\frac{a+\epsilon}{2}$$

Since $\epsilon$ was arbitrary we conclude

$$\frac{a}{2}\leq \lim_{n\to\infty}b_n\leq \frac{a}{2}$$

We conclude

$$\lim_{n\to\infty}b_n=\frac{a}{2}$$

EDIT: As requested, if for all $\epsilon>0$ we have

$$a_n-\epsilon<b_n<a_n+\epsilon$$

and $a_n\to a$, then $b_n\to a$. First, note that we can rearrange this to

$$|b_n-a_n|<\epsilon$$

Let $\delta>0$ be given and choose $\epsilon=\frac{\delta}{2}$. Then choose $N$ such that $n\geq N$ implies $|a_n-a|<\frac{\delta}{2}$. This gives us

$$|b_n-a|=|b_n-a+a_n-a_n|\leq |b_n-a_n|+|a-a_n|<\frac{\delta}{2}+\frac{\delta}{2}=\delta$$

as desired.

Solution 2:

An application of Stolz-Cesaro works: \begin{align} \lim_{n \to \infty} \frac{\sum_{i=1}^n ia_i}{n^2} = \frac{a}{2} &\color{red}\impliedby \lim_{n \to \infty} \frac{\sum_{i=1}^{n+1} ia_i - \sum_{i=1}^n ia_i}{(n+1)^2 - n^2} = \frac{a}{2} \\ &\iff \lim_{n \to \infty} \frac{(n+1)a_n}{2n + 1} = \frac{a}{2} \\ &\impliedby \lim_{n \to \infty} \frac{n+1}{2n + 1} \cdot \lim_{n \to \infty} a_n = \frac{a}{2} \\ &\iff \frac{1}{2} \cdot \lim_{n \to \infty} a_n = \frac{a}{2}, \end{align} which is true. The red $\color{red}\impliedby$ marks where Stolz-Cesaro was used. Importantly, note that $n^2$ diverges to $\infty$ strictly monotonically, so the theorem applies.

Solution 3:

We rely on the following lemma:

Lemma (Stolz-Césaro): Suppose $y_n\ge0$, $x_n=o(y_n)$, and $\sum_{k\le n}y_k\to+\infty$ as $n\to+\infty$. Then

$$ \sum_{k\le n}x_k=o\left(\sum_{k\le n}y_k\right) $$

Before go onto proving the lemma, let's see how it helps to solve OP's problem:

Since $n^2\sim n(n+1)$ as $n\to+\infty$, we can plug in $x_n=n(a_n-a)$ and $y_n=2n$. It is easy to show that $x_n=o(y_n)$, so we can now apply the lemma to obtain

$$ \lim_{n\to\infty}{1\over n^2}\sum_{k\le n}k(a_k-a)=0 $$

This clearly indicates that

$$ \lim_{n\to\infty}{1\over n^2}\sum_{k\le n}ka_k=\frac a2 $$

Proof of the lemma.

By definition, we know for all $\varepsilon>0$ there exists $N=N(\varepsilon)$ such that $|x_n|<\varepsilon y_n$ holds for all $n>N$. This means

$$ \left|\sum_{k\le n}x_k\right|<\left|\sum_{k\le N}x_k\right|+\varepsilon\sum_{N<k\le n}y_k $$

As a result, we see that for all $\varepsilon>0$

$$ \limsup_{n\to\infty}\left|\sum_{k\le n}x_k\right|\left(\sum_{k\le n}y_k\right)^{-1}<\varepsilon $$

Consequently the limit is zero, concluding the proof.

Solution 4:

For any $\varepsilon > 0$ we have $N$ such that any $n \ge N$, $a - \varepsilon \le a_n \le a + \varepsilon$ holds. Let $M_1 = \sum_{i = 1}^{N}{i(a - \varepsilon)}$ and $M_2 = \sum_{i = 1}^{N}{i(a + \varepsilon)}$ which is obviously finite and fixed. Notice that $$ \frac{M_1}{n^2} + \frac{1}{n^2}\sum_{i = N+1}^{n}{i(a - \varepsilon)} \le b_n \le \frac{M_2}{n^2} + \frac{1}{n^2}\sum_{i = N+1}^{n}{i(a + \varepsilon)}$$ Simplify this inequality and we get $$ \frac{M_1}{n^2} + \frac{a - \varepsilon}{2}\left(1 - \frac{N}{n}\right)\left(1 + \frac{N + 1}{n}\right) \le b_n \le \frac{M_2}{n^2} + \frac{a + \varepsilon}{2}\left(1 - \frac{N}{n}\right)\left(1 + \frac{N + 1}{n}\right)$$ Take $n\to\infty$ then $(a - \varepsilon)/2 \le \lim_{n\to\infty}{b_n} \le (a + \varepsilon)/2$. Note that this holds for any $\varepsilon > 0$ so this proves what you want.