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^p}, \end{align*} where $1 < p < 2$. The sequence is clearly Cauchy and convergent. Let $x$ be the limit point. I am wondering whether we could get some convergence rate inequality \begin{align*} \|x_k - x\| \le f(k), \end{align*} where $f(k)$ is some function of $k$.
We clearly can do this for $p=2$. That is $\|x_k -x\| \le \frac 1 k$ by telescoping. The answer is in the question I asked before (actually comments in the accepted answer). But I am not sure if we can still use the telescoping technique.
$$\|x_k-x\|=\left\|\sum_{i=k}^\infty x_i-x_{i+1}\right\|\leq\sum_{i=k}^\infty\|x_i-x_{i+1}\|\leq\sum_{i=k}^\infty\frac{1}{i^p}.$$ The last series is most easily estimated using an integral: $$\sum_{i=k}^\infty\frac{1}{i^p}\leq\sum_{i=k}^\infty\int_{i-1}^i\frac{\mathrm dx}{x^p}=\int_{k-1}^\infty\frac{\mathrm dx}{x^p}=\frac{1}{p-1}\frac{1}{(k-1)^{p-1}}.$$