Question

For each n let an = (1+ 1/n)ⁿ

Prove that an is convergent as follows:
1) Show its increasing
2) show its bounded

Answer 1

To prove that the sequence an = (1 + 1/n)ⁿ is convergent, we need to show that it is increasing and bounded. We can show that the sequence is increasing by taking the ratio of consecutive terms and showing that it is greater than 1. To show that the sequence is bounded, we can use the binomial theorem to expand (1 + 1/n)ⁿ and find an upper bound for the terms.

Step-by-step explanation:

To prove that the sequence an = (1 + 1/n)ⁿ is convergent, we need to show that it is increasing and bounded.

1. Show that the sequence is increasing: We can prove that the sequence is increasing by showing that consecutive terms are greater than the previous ones. To do this, we can take the ratio of an+1 and an, and show that it is greater than 1. By simplifying the ratio, we get (1 + 1/n+1)^n+1 / (1 + 1/n)^n. Applying some algebraic manipulations and simplifying the expression, we get (1 + 1/n+1) * (1 + 1/n)^n > 1. Since (1 + 1/n)^n > 0 for all n, and (1 + 1/n+1) > 1, the inequality (1 + 1/n+1) * (1 + 1/n)^n > 1 holds, which proves that the sequence is increasing.
2. Show that the sequence is bounded: To show that the sequence is bounded, we need to find an upper bound M such that an < M for all n. To do this, we can use the binomial theorem to expand (1 + 1/n)ⁿ. By expanding and simplifying the expression, we get an = 1 + n(1/n) + n(n-1)(1/n)²/2! + ..., which can be further simplified to an = 1 + 1 + (n-1)/2! + (n-1)(n-2)/3! + ... + (n-1)(n-2)...1/n!. Since each term in the expansion is positive, it follows that an < 1 + 1 + 1/2! + 1/3! + ... + 1/n!, which is a finite sum. Therefore, we can take M = 1 + 1 + 1/2! + 1/3! + ... + 1/n! as the upper bound for the sequence an, which proves that the sequence is bounded.