MATH 295 PROBLEMS 4

IGOR KRIZ

1. Compute from first principles:

lim_{n→∞}(\sqrt{n + 1} − \sqrt{n}).

[Hint: Use (a − b)(a + b) = a^{2} − b^{2}.]

2. A sequence is given as follows: a_{1} = \frac{1}{2}, a_{n+1} = \frac{1+a_{n}}{2} . Decide if this sequence converges, and if it does, compute its limit.

3. Let a_{1}, ..., a_{n} ≥ 0. Prove that the arithmetic mean

A_{n} = \frac{a_{1} + ... + a_{n}}{n}

and the geometric mean

G_{n} = \sqrt[n]{a_{1}...a_{n}}

satisfy

G_{n} ≤ A_{n}.

When does equality hold? [Hint - one possible way: First prove G_{n} ≤ A_{n} for n = 2. By induction on k, then prove that G_{n} ≤ A_{n} for n = 2^{k}. For a general n, let 2^{k} > n. Consider the 2^{k} numbers

a_{1}, ..., a_{n}, \underbrace{A_{n}, ..., A_{n}}_{2^{k} − n times}

to prove G_{n} ≤ A_{n}.]

4. Suppose (a_{n}) is a sequence of positive numbers such that

lim_{n→∞} a_{n} = 0.

Let b_{n} = a_{1} − a_{2} + a_{3} − ... + (−1)^{n+1}a_{n}. Prove that then the sequence b_{n} converges. [Comment: (−1)^{n+1} is +1 if n is odd and −1 if n is even. We use this expression in the above formula simply to get the right sign for a_{n}.]

5. Prove that every sequence (a_{n}) such that

|a_{n} − a_{n+1}| < \frac{1}{2^{n}} for every n ∈ ℕ

converges.