Question

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{c n}{b n+1}=\frac{c}{b}, \text { for real numbers } b > 0 \text { and } c > 0$$

Short answer

Question: Prove that the limit of the sequence $$\frac{cn}{bn+1}$$ as n approaches infinity is $$\frac{c}{b}$$ using the formal definition of a limit. Answer: We can prove the limit by selecting an epsilon (ε) and finding a positive integer (N) such that for all integers n > N, the absolute difference between the sequence and the expected limit is less than ε. We manipulate the inequality $$\left|\frac{cn}{bn+1} - \frac{c}{b}\right| < \epsilon$$ and find a suitable N by taking the smallest integer value greater than the obtained inequality $$N = \left\lceil\frac{c}{b^2\epsilon} - \frac{1}{b}\right\rceil$$. This confirms the limit $$\lim _{n \rightarrow \infty} \frac{c n}{b n+1}=\frac{c}{b}$$.

Step-by-step solution

Set up the inequality

We want to find a positive integer N such that for all integers n > N, the following inequality holds true: $$\left|\frac{cn}{bn+1} - \frac{c}{b}\right| < \epsilon$$

Manipulate the inequality

Combine the fractions to work with a single fraction. $$\left|\frac{bcn - c(bn+1)}{b(bn+1)}\right| < \epsilon$$ Simplify the numerator and factor c. $$\left|\frac{c(-1)}{b(bn+1)}\right| = \frac{c}{b(bn+1)} < \epsilon$$ As c, b > 0, the fraction will always be positive, so we can drop the absolute value. $$\frac{c}{b(bn+1)} < \epsilon$$

Set up an inequality for N

To make the left side of the inequality less than ε, we can use the following observation: $$\frac{c}{b(bn+1)} < \frac{c}{b(bN)}$$ Therefore, $$\frac{c}{b(bN)} < \epsilon$$ From this, we can solve for N. $$N > \frac{c}{b^2\epsilon} - \frac{1}{b}$$

Define N

We can take the smallest integer value greater than the obtained inequality for a valid N: $$N = \left\lceil\frac{c}{b^2\epsilon} - \frac{1}{b}\right\rceil$$

Confirm the result

For any ε, we have now found an integer N such that if n > N, then $$\left|\frac{cn}{bn+1} - \frac{c}{b}\right| < \epsilon$$ This confirms the limit: $$\lim _{n \rightarrow \infty} \frac{c n}{b n+1}=\frac{c}{b}$$