1. The sequence is increasing and bounded above by 5, so it converges. The limit of the terms is 5

1. Each term after a1 is the midpoint of 5 and the previous term, so the sequence is increasing and bounded above by 5. It follows from the Monotone Convergence Theorem that this sequence converges. Now that we know the limit lim_{n → ∞}a_n exists, denote this limit by α. Taking the n → ∞ limit of both sides of the equation: a_n= (a_n-1 + 5)/2 we find: α= (α + 5)/2 Solving for α gives α= 5.

1. Convergence follows from two applications of the Monotone Convergence Theorem, and an estimate on the difference of successive terms. The sequence converges to (-1 + √5)/2.

1. The first few values of the ratios b_n= a_n/a_n+1 are 1, .5, .6667, .6000, .6250, .6154, .6190, .6176, .6182, .6180, .6181, .6180, .6180, … . It appears that the sequence does converge. Now assuming the limit lim_{n → ∞}a_n/a_n+1 exists, denote this limit by α. Divide both sides of the equation a_n= a_n-1 + a_n-2 by a_n-1, obtaining a_n/a_n-1= 1 + a_n-2/a_n-1 Taking the n → ∞ limit of both sides of this equation gives 1/α= 1 + α Multiplying both sides by α gives a quadratic equation with solutions α= (-1 ± √5)/2. Being the limit of ratios of positive terms, α cannot be negative, so α= (-1 + √5)/2

1. The series diverges.

1. This is easy if you recall the quotient property of logarithms log(n/(n+1)) = log(n) – log(n+1) So the series telescopes (log(1) – log(2)) + (log(2) – log(3)) + (log(3) – log(4)) + … so the n^th partial sum is Sn = log(1) – log(n+1) = -log(n+1), which diverges to -∞.

1. The series diverges.

1. Let’s look for a pattern in the first few terms: a₁= 1/2 a₂= 1/2² + 1/2 = (1 + 2)/2² a₃= 1/2³ + (1 + 2)/2² = (1 + 2 + 2²)/2³ and in general a_n = (1 + 2 + … + 2ⁿ)/2ⁿ⁺¹ Now recall 1 + 2 + 2² + … + 2ⁿ= 2ⁿ⁺¹ – 1. (Easy to prove by induction if you don’t recall this result.) So we see that: a_n= (2ⁿ⁺¹ – 1)/2ⁿ⁺¹ and the lim_{n → ∞}a_n= 1. The series diverges by the n^th term test.