Question

Let {bn} be a sequence such that bn = 1 + (1/n). Prove that inf(bn) = 1

Answer 1

Answer on Question #74429 – Math – Real Analysis

Question

Let $\{b_n\}$ be a sequence such that $b_n = 1 + \frac{1}{n}$. Prove that $\inf(b_n) = 1$

Solution

1) Since $\frac{1}{n} > 0$ all $b_n > 1$ and hence $\inf(b_n) \geq 1$

2) For any $\epsilon > 0$ let us take $N_{\epsilon} = \left[\frac{1}{\epsilon}\right] + 1$. Since $N_{\epsilon} > \frac{1}{\epsilon}$, $\frac{1}{N_{\epsilon}} < \epsilon$ and $b_{N_{\epsilon}} < 1 + \epsilon$. Hence for any $\epsilon > 0$, $\inf(b_n) \leq b_{N_{\epsilon}} < 1 + \epsilon$. Therefore $\inf(b_n) \leq 1$.

Since $\inf(b_n) \geq 1$ and $\inf(b_n) \leq 1$, $\inf(b_n) = 1$.

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