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Suppose $\{x_k\} \subseteq X$ where $X$ is a Banach space. $\{x_k\}$ satisfies \begin{align*} \|x_k - x_{k-1} \| \le \frac 1 {k^2}. \end{align*} It is clear $\{x_k\}$ is convergent since for sufficiently large $n, m$, \begin{align*} \|x_n - x_m\| \le \sum_{j=n}^m \frac 1 {j^2} \le \varepsilon. \end{align*} I would like to know whether it makes sense to consider the limit $\lim_{k \to \infty} \left( k x_k - (k-1) x_{k-1} \right)$. If it makes sense, is the computation following correct? \begin{align*} \lim_{k \to \infty} \left( k(x_k - x_{k-1}) + x_{k-1} \right) = \lim_{k \to \infty} k(x_k - x_{k-1}) + \lim_{k \to \infty} x_{k-1} = x \\ \end{align*}

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Yes, in any normed space, if $x_n \to a$ and $y_n \to b$ then $x_n+y_n \to a+b$.

Also, note that $$\| k(x_k - x_{k-1}) \| = |k| \|x_k - x_{k-1} \| \leq \frac{1}{k}$$ implies $$\lim_{k \to \infty} (k(x_k - x_{k-1})) =0 $$

In general $x_n \to a$ in a normed space if and only if $\|x_n-a \| \to 0$ in $\mathbb R$.

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  • $\begingroup$ Thanks for your answer. Is it possible to estimate how fast $x_n \to a$ as a function of $k$ in this scenario? $\endgroup$ – user1101010 Jan 23 '18 at 22:47
  • $\begingroup$ @jing007 Yes, $$\|x_n -a \| \leq \sum_{k=n}^\infty \| x_k -x_{k+1} \| \leq \sum_{k=n}^\infty \frac{1}{(k+1)^2} \leq \sum_{k=n}^\infty \frac{1}{k(k+1)}=\frac{1}{n} $$ since the last series is telescopic. $\endgroup$ – N. S. Jan 23 '18 at 22:52
  • $\begingroup$ @jing007 If you need more details for any step in the computation, let me know. $\endgroup$ – N. S. Jan 23 '18 at 23:37

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