$(X,d)$ be a complete metric space, and $d(x_{n},x_{n+1}) \leq \frac{1}{n^2} , d(y_{n},y_{n+1}) \leq \frac{1}{n}$ .

Question

Let $(X,d)$ be a complete metric space, and $x_{n} , y_{n}$ be sequences in $X$ such that $d(x_{n},x_{n+1}) \leq \frac{1}{n^2} , d(y_{n},y_{n+1}) \leq \frac{1}{n}$

then does $(x_{n}) , (y_{n})$ sequence converge?

Suppose they converge, that is $x_{n} \rightarrow x , y_{n} \rightarrow y$ and since $X$ is complete so $x,y \in X$.

Using triangle inequality I got $d(x_{n},x) \leq \sum \frac{1}{n^2} = \frac{\pi^2}{6}$ but I cannot also say that they tend to zero?

and also $d(y_{n},y) \leq \sum \frac{1}{n}$

How taking Cauchy sequence will help?

But i also think that $\frac{1}{n^2} \leq \frac{1}{n}$.So we also have $d(x_{n},x) \leq \sum \frac{1}{n^2} \leq \sum \frac{1}{n}$.

Answer 1

$d(x_{m},x_{n})\leq d(x_{m},x_{m-1})+\cdots+d(x_{n+1},x_{n})\leq\dfrac{1}{m^{2}}+\cdots+\dfrac{1}{n^{2}}<\displaystyle\sum_{k\geq n}\dfrac{1}{k^{2}}$ for $m>n$. As $\displaystyle\sum_{k\geq n}\dfrac{1}{k^{2}}\rightarrow 0$, $n\rightarrow\infty$, so $\{x_{n}\}$ is Cauchy and hence converges by the completeness.

Answer 2

For the case of $x$, you wanna prove the sequence is Cauchy. Take $n, m$, and suppose that $n < m$. Then \begin{align*} d(x_n , x_m) & \leq d(x_n , x_{n + 1}) + \cdots + d(x_{m - 1} , x_m) \\ & \leq n^{-2} + \cdots + (m - 1)^{-2} \\ & \leq \frac{1}{n^2} + \frac{1}{(n + 1)^2} + \cdots . \end{align*} This last expression goes to $0$ as $n \to \infty$.

As for $y$, you can make a sequence that satisfies the condition but doesn't converge. Let $y_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n - 1}$.