Prove that if $a_n \rightarrow L$ then the sequence $b_n = 1 / n^2 \cdot (a_1 + 2a_2 + 3a_3 + ... + na_n)$ converges to $0.5L$ As the title says.
I assumed that $c_n = \frac{1}{n^2}$ and $a_n = \sum_{i=1}^ni\cdot a_i \rightarrow L$.
Tried to use the limit properties but all I got was $lim(1) \cdot lim(L)$ which is not as expected(raised a concern that "n" may has something to do with the positioning of each element in $a_n$?).
 A: Here is a direct argument.
Let $\epsilon>0$ be arbitrary, then find $N\in \mathbb{N}$ so that $n  \geq N$ implies $L-2\epsilon <a_n<L+2\epsilon$ for $n \geq N$.
This implies, for $n\geq N$, that $$\sum_{k=1}^{N-1}ka_k+\sum_{k=N}^nk(L-2\epsilon)<\sum_{k=1}^nka_k<\sum_{k=1}^{N-1}ka_k+\sum_{k=N}^nk(L+2\epsilon)$$ We can write the above inequalities as $$\Bigg[\sum_{k=1}^{N-1}ka_k-\sum_{k=1}^{N-1}k(L-2\epsilon)\Bigg]+\sum_{k=1}^nk(L-2\epsilon)<\sum_{k=1}^nka_k<\Bigg[\sum_{k=1}^{N-1}ka_k-\sum_{k=1}^{N-1}k(L+2\epsilon)\Bigg]+\sum_{k=1}^nk(L+2\epsilon)$$ Divide by $n^2$:  $$\frac{1}{n^2}\Bigg[\sum_{k=1}^{N-1}ka_k-\sum_{k=1}^{N-1}k(L-2\epsilon)\Bigg]+\sum_{k=1}^n\frac{1}{n}\cdot \frac{k}{n}(L-2\epsilon)<\frac{1}{n^2}\sum_{k=1}^nka_k<\frac{1}{n^2}\Bigg[\sum_{k=1}^{N-1}ka_k-\sum_{k=1}^{N-1}k(L+2\epsilon)\Bigg]+\sum_{k=1}^n\frac{1}{n}\cdot \frac{k}{n}(L+2\epsilon)$$ Take $n\longrightarrow \infty$ and use squeeze theorem to deduce $$\int_0^1(L-2\epsilon)x\mathrm{d}x \leq \lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n ka_k \leq \int_0^1 (L+2\epsilon)x\mathrm{d}x$$ This is equivalent to $$0.5L- \epsilon \leq \lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n ka_k \leq 0.5L + \epsilon$$ The result follows since $\epsilon>0$ is arbitrary.
