Convergence implies boundness of a sequence but an example confuses me! given the following sequence
$$
p_{n} = \frac{1}{n-1}
$$
I am tempted to say that for n=1 the sequence assumes an infinite (and thus unbounded) value but still the sequence converges to zero as n increases.
I am of course wrong because this contradicts theorem 3.2c of Baby Rudin which states that &quot;if a sequence is convergent in a metric space then it's bounded&quot;. What am I missing? Maybe 1/0 is not a mathematical defined value?
I would be greatly thankful for an answer.
Erma
Single values for which $p_n$ is not defined are not relevant when we take the limit as $n\to \infty$ what matters is that $p_n$ is eventually a well defined expression.
For example
$$\lim_{n\to \infty} \frac1{\sqrt{n^2-1\,000\,n-1\,000\,000}}$$
is not well defined for a finite number of values but the limit is equal to $0$ since eventually the expression inside the square root becomes positive and tends to $\infty$.
Single values for which $p_n$ is not defined are not relevant when we take the limit as $n\to \infty$ what matters is that $p_n$ is eventually a well defined expression.
For example
$$\lim_{n\to \infty} \frac1{\sqrt{n^2-1\,000\,n-1\,000\,000}}$$
is not well defined for a finite number of values but the limit is equal to $0$ since eventually the expression inside the square root becomes positive and tends to $\infty$.