Find the limit of the sequence $\{\frac{x_1 + 2x_2 + ... + nx_n}{n^2}\}$, where $\{x_n\}$ is a convergent sequence in a normed space S.

Question

Let $\{x_n\}_{n=1}^{\infty}$ be a convergent sequence in a normed space $(S, ||\cdot ||)$. Find the limit of the sequence \begin{align} \left\{\frac{x_1 + 2x_2 + \cdots + nx_n}{n^2}\right\}_{n=1}^{\infty} . \end{align}

Could you please give me some hints where to start?

Answer 1

You can use the Stolz–Cesàro theorem, which is a l'Hôpital's rule for sequences.

Let $a_n = \sum_{i=1}^n ix_i$, and $b_n=n^2$. We have $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{(n+1)x_{n+1}}{2(n+1)-1} \rightarrow \frac{x}{2}, when \ \ n\rightarrow \infty$$ so $\frac{a_n}{b_n}$ converges to $\frac{x}{2}$.

Of course $\delta-\epsilon$ arguments should work here too.

Answer 2

This looks a bit like Cesáro mean taken a level higher.

Let $x=\lim x_n$. First, guess the correct limit in terms of $x$ - for example by considering the case of a constant given sequence $\{x_n\}$. Then, make and $\epsilon$-proof: For given $\epsilon>0$, we have $kx_k\approx kx$ for $k$ big enough. On the other hand the contribution of the first summands gets negligible as $n\to \infty$ as it is divided by $n^2$.

I suppose you can fill in the details from these hints.

Answer 3

For $\varepsilon > 0$ choose $N>0$ such that $n>N \Longrightarrow |x_n-x|<\varepsilon$. Now, since $\frac{1+2+\dots+n}{n^2} = \frac{1}{2} + \frac{1}{2n}$ \begin{align} \left| \frac{x_1+2x_2+\dots+nx_n}{n^2} - \frac{x}{2} \right|&= \left| \frac{x_1+2x_2+\dots+nx_n}{n^2} - \left(\frac{1+2+\dots+n}{n^2} - \frac{1}{2n} \right)x \right|\\ & = \left| \frac{1}{n^2}\sum_{k=1}^{n}k(x_k-x) + \frac{x}{2n} \right| \\ & = \left| \frac{1}{n^2}\sum_{k=1}^{N}k(x_k-x) + \frac{1}{n^2}\sum_{k=N+1}^{n}k(x_k-x) + \frac{x}{2n} \right| \\ & \leqslant \left| \frac{1}{n^2}\sum_{k=1}^{N}k(x_k-x) \right| + \left| \frac{1}{n^2}\sum_{k=N+1}^{n}k(x_k-x) \right| + \left| \frac{x}{2n} \right| \\ & \leqslant \frac{1}{n^2}\sum_{k=1}^{N}\left|k(x_k-x)\right| + \frac{1}{n^2}\sum_{k=N+1}^{n}\left|k(x_k-x) \right| + \left| \frac{x}{2n} \right| \\ & \leqslant \frac{1}{n^2}\sum_{k=1}^{N}\left|k(x_k-x)\right| + \frac{1}{n^2}\sum_{k=N+1}^{n}k \varepsilon + \left| \frac{x}{2n} \right| \\ & \leqslant 3 \varepsilon \end{align}

for $n$ sufficiently large.