Find the supremum least upper bound infimum greatest lower

Question

Exercise 1 Prove by induction:

(a) 1² + 2² +..........+ n² = n(n + 1)(2n + 1)/6, ∀n ∈ N.

(b) 1³ + 2³ + ........+ n³ = {n(n + 1)/2}², ∀n ∈ N.

(c) Σⁿ_k=1 k(k + 1) = n(n + 1)(n + 2), ∀n ∈ N.

(d) 1 + 3 + 5 + .....+ (2n - 1) = n², ∀n ∈ N.

(e) 3 + 11 + 19 + 27 + .....+ (8n - 5) = 4n² - n, ∀n ∈ N.

(f) 1 + 2 + 3 +.......+ (2n) = 2n² + n, ∀n ∈ N.

Exercise 2 Prove by induction:

(a) Σⁿ_k=1 k.k! = (n + 1)! - 1, ∀n ∈ N.

(b) 1/1.2 + 1/2.3 + 1/3.4 +.........+ 1/n(n + 1) = n/n+1, ∀n ∈ N.

(c) 1/1.3 + 1/3.5 + 1/5.7 +.........+ 1/(2n - 1)(2n + 1) = n/n+1, ∀n ∈ N.

Exercise 3 Prove by induction the following equalities:

(a) (1 - 1/2).(1 - 1/3).........(1-1/n) = 1/n, ∀n ∈ N, n ≥ 2.

(b) (1 - 1/2²).(1 - 1/3²).........(1-1/n²) = (n + 1)/2.n, ∀n ∈ N, n ≥ 2.

Exercise 4 Prove by induction:

(a) 1 + r + r² + r³ + ........+ rⁿ = (rⁿ⁺¹ -1)/(r - 1) if r ∈ R, r ≠ 1, ∀n ∈ N.

(b) 1 + 1/2 + 1/2² +...........1/2ⁿ, ∀n ∈ N.

Exercise 5 Prove by induction:

(a) 7ⁿ - 2ⁿ is a multiple of 5, ∀n ∈ Z, n ≥ 0.

(b) 3⁴ⁿ + 9 is a multiple of 10, ∀n ∈ Z, n ≥ 0.

Exercise 6 Prove by induction the following inequalities:

(a) 2ⁿ < n!, ∀n ∈ N, n ≥ 4.

(b) n² < 2ⁿ, ∀n ∈ N, n ≥ 5.

(c) n² < n!, ∀n ∈ N, n ≥ 4.

(d) 1 + 1/2 + 1/3 + 1/4 + 1/5 +.........+ 1/2ⁿ ≥ 1 + n/2, ∀n ∈ N.

Exercise 7 Prove that for all a, b > 0,

2ab/(a + b) ≤ √ab ≤ (a + b)/2,

that is, the harmonic mean, H = 2ab/(a + b), is less or equal than the geometric mean, G = √ab, and also this later is less or equal than the arithmetic mean, A = (a + b) /2.

Exercise 8 Find all real numbers a ≥ -1 such that
a < √(1 + a/2)

Exercise 9 Find the supremum (least upper bound), infimum (greatest lower bound), maximum and minimum of the following sets, if they exist.

(a) A = { 1/n : n ∈ N}

(b) B = { 1/n +(-1)ⁿ : n ∈ N}

(c) C = {x : 0 ≤ x ≤ √2 and x rational }.

(d) D = {x : x² + 5x - 6 < 0}.

Exercise 10 Find the real numbers x that satisfy the following conditions.

(a) |x + 1| < 2                 (b) |x + 1/x-1|< 2

(c) |x - 1|.|x + 1| = 0       (d) |x - 1| + |x - 2| > 1

(e) |x - 1| + |x + 1| < 1    (f) |x - 1|.|x + 2| = 3

(g) |x - 1|.|x + 2| = 1       (h) |2x - 1| < |x - 1|

(i) |x .(1 - x)| < 1