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Let $x_n$ and $y_n$ be two sequences. Given that $\left| x_n - y_n \right|<\frac{1}{n}$ and $x_n \to \pi$, prove that $b_n\to \pi$.

My approach:

Since $x_n$ converges as $n\to\infty$, $y_n$ must also converge. Suppose that $y_n\to c$ as $n\to \infty$. Let $\epsilon_y := \frac{1}{n}$, then $y_n < \left|\frac{1}{n}+c\right|$. We also know that $\forall \varepsilon_x > 0$, $\exists N_{\varepsilon_x} > 0$ such that $n>N_{\varepsilon_x}$ implies $\left|x_n - \pi\right|<\varepsilon_x$.

Now, let $\varepsilon:= 1$, then $\left| x_n-y_n \right|<\frac{1}{n}<1=\varepsilon_x$. Thus $x_n<\left|\frac{1}{n}+y_n\right|<\left|\frac{2}{n}\right|+c$. Let $n\to \infty$, then $x_n\to c$. Therefore, $c=\pi$.

Please let me know if this proof is OK, and is it too complicated?

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Your proof has several potentially false inequalities. For example, suppose $x_n = \pi + \frac{5}{n}$ and $y_n = \pi$ for all $n$. Is it true that $x_n < \left|\frac{1}{n} + y_n\right|$? Is it true that $|x_n - y_n| < \frac{1}{n}$?

As a hint: $|y_n - \pi| \leq |y_n - x_n| + |x_n - \pi|$ for all $n$ by the triangle inequality.

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