# Prove that if a n </msub> &#x2192;<!-- → --> L then the sequence b n

Bailee Landry 2022-05-24 Answered
Prove that if ${a}_{n}\to L$ then the sequence ${b}_{n}=1/{n}^{2}\cdot \left({a}_{1}+2{a}_{2}+3{a}_{3}+...+n{a}_{n}\right)$ converges to 0.5L
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Let $ϵ>0$ be arbitrary, then find $N\in \mathbb{N}$ so that $n\ge N$ implies $L-2ϵ<{a}_{n} for $n\ge N$
This implies, for $n\ge N$ , that
$\sum _{k=1}^{N-1}k{a}_{k}+\sum _{k=N}^{n}k\left(L-2ϵ\right)<\sum _{k=1}^{n}k{a}_{k}<\sum _{k=1}^{N-1}k{a}_{k}+\sum _{k=N}^{n}k\left(L+2ϵ\right)$
We can write the above inequalities as
$\left[\sum _{k=1}^{N-1}k{a}_{k}-\sum _{k=1}^{N-1}k\left(L-2ϵ\right)\right]+\sum _{k=1}^{n}k\left(L-2ϵ\right)<\sum _{k=1}^{n}k{a}_{k}<\left[\sum _{k=1}^{N-1}k{a}_{k}-\sum _{k=1}^{N-1}k\left(L+2ϵ\right)\right]+\sum _{k=1}^{n}k\left(L+2ϵ\right)$
Divide by ${n}^{2}$
$\frac{1}{{n}^{2}}\left[\sum _{k=1}^{N-1}k{a}_{k}-\sum _{k=1}^{N-1}k\left(L-2ϵ\right)\right]+\sum _{k=1}^{n}\frac{1}{n}\cdot \frac{k}{n}\left(L-2ϵ\right)<\frac{1}{{n}^{2}}\sum _{k=1}^{n}k{a}_{k}<\frac{1}{{n}^{2}}\left[\sum _{k=1}^{N-1}k{a}_{k}-\sum _{k=1}^{N-1}k\left(L+2ϵ\right)\right]+\sum _{k=1}^{n}\frac{1}{n}\cdot \frac{k}{n}\left(L+2ϵ\right)$
Take $n⟶\mathrm{\infty }$ and use squeeze theorem to deduce
${\int }_{0}^{1}\left(L-2ϵ\right)x\mathrm{d}x\le \underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=1}^{n}k{a}_{k}\le {\int }_{0}^{1}\left(L+2ϵ\right)x\mathrm{d}x$
This is equivalent to
$0.5L-ϵ\le \underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=1}^{n}k{a}_{k}\le 0.5L+ϵ$
The result follows since $ϵ>0$ is arbitrary.