Find limit of sum $\lim_{n\to \infty}\frac{1}{n^3}\sum^{n}_{k=1}k^2a_k$ If $\lim_{n\to \infty}a_n = A$, find $$\lim_{n\to \infty}\left(\frac{1}{n^3}\sum^{n}_{k=1}k^2a_k\right)$$
Any hints? I thought that I can interchange $\lim$ with sum, but I think that it is not true, because I don't know type of convergence of sum. Also, I know that $\sum a_k$ should converge because $\sum^{n}_{k=1}k^2a_k$ exists. But how to proceed from here?
 A: You can interchange $a_k$ with its limit $A$,
$$\lim_{n\to\infty}\left|\frac{1}{n^3}\sum_{k=1}^nk^2 a_k - \frac{1}{n^3}\sum_{k=1}^nk^2A\right| = 0.$$
To see this, let $\varepsilon>0$. First note, that $|a_k|\le C$ for some $C$. Moreover, let $K>0$ such that $|a_k-A|\le \varepsilon$ for all $k\ge K$. Then, for $n\ge K$ there holds
\begin{align*}
\left|\frac{1}{n^3}\sum_{k=1}^nk^2 a_k - \frac{1}{n^3}\sum_{k=1}^nk^2A\right| & \le \frac{1}{n^3}\sum_{k=1}^N k^2|(a_k-A)| + \frac{1}{n^3}\sum_{k=N+1}^n k^2|a_k-A|\\
&\le \frac{K^3(|A|+C)}{n^3} + \frac{1}{n^3}\sum_{k=1}^n k^2 \varepsilon \le \frac{K^3(|A|+C)}{n^3} + \varepsilon \le 2\varepsilon
\end{align*}
for $n$ sufficiently large. Therefore the limit is $\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^nk^2 a_k = A/3$.
A: Denote the partial sums with $s_n = \sum^{n}_{k=1}k^2a_k$ and apply the Stolz–Cesàro theorem:
$$
 \lim_{n \to \infty} \frac{s_n}{n^3} =  \lim_{n \to \infty} \frac{s_n - s_{n-1}}{n^3 - (n-1)^3}
$$
holds if the limit on the right exists, which is the case:
$$
 \frac{s_n - s_{n-1}}{n^3 - (n-1)^3} =  \frac{n^2 a_n}{n^3 - (n-1)^3}
=  \frac{n^2 }{3 n^2 - 3n +1}a_n \to \frac 13 A
$$
for $n \to \infty$.
This is also valid if the sequence $(a_n)$ diverges to $+\infty$ or to $-\infty$.
A: There is a general theorem:

If $b_n$ is a sequence and $b_n\to L.$ Let: $$c_n=\frac{b_1+\cdots+b_n}{n}.$$
Then $c_n\to L.$

In this problem, let $b_n$ be the sequence:
$$a_1,a_2,a_2,a_2,a_2,\underbrace{a_3,\dots,a_3}_{9 \text{ times}},\dots,  \underbrace{a_k,\dots,a_k}_{k^2\text{ times}},\dots$$
Then we get, from the theorem, that:
$$\frac{\sum_{k=1}^nk^2a_k}{\sum_{k=1}^nk^2}\to A.$$
Then you need to compute the limit:
$$\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n k^2$$

The theorem above is used in Cesàro summation, and it is a special case of the Stolz-Cesàro theorem, used in another answer.
