Solution 4

1. If a, b are both ≥ 0 or both ≤ 0 then we have equality. If (say) a > 0, b < 0 so that -b = c > 0 then the LHS = a - c and the RHS = a + c and -c < c. Similarly if a < 0, b > 0.
   
   It is called the triangle inequality since if a, b ∈ C and | | denotes the modulus of a complex number, this is the geometric theorem that the sum of the lengths of any two sides of a triangle is ≥ the length of the third side.
   
   The above proves the result for n = 2. Assume that it is true for n = k and consider it for n = k + 1.
   |a₁+ a₂+ ... + a_k+ a_k+1| ≤ |a₁+ a₂+ ... + a_k| + |a_k+1| by the n = 2 case and this is ≤ (|a₁| + |a₂| + ... + |a_k|) + |a_k+1| by the inductive assumption.
   
   

2. (a_n)→ α and (b_n)→ α. Since a_n≤ x_n≤ b_n we have a_n- α ≤ x_n- α ≤ b_n-α. Now, given ε > 0, when n is large enough, a_n- α and b_n- α both lie in an interval (-ε, ε) around 0 and since x_n- α is between them it also lies in this interval. This is the condition for convergence.

3. (a) (1, ^-1/₂, ¹/₃, ^-1/₄, ... )→ 0 but is not monotonic.
   (b) (1, 0, 1, 0, ... ) is bounded but not convergent.
   (c) Every convergent sequence is bounded.
   (d) (1, 2, 3, ... ) is monotonic (increasing) but not bounded.
   (e) (1, 1², 2, 2², 3, 3², ... ) diverges to ∞ but not monotonically.
   (f) Monotonic bounded sequences are always convergent.
   (g) (1, -1, 2, -2, ... ) is unbonded and not monotonic.
   (h) (¹/₂, ¹/_2²,¹/₃, ¹/_3²,... )→ ∞ but is not monotonic.

4. We must choose N so that |(2n + 1)/(n + 1) - 2| < ε for n > N. That is 2/(n + 1) < ε or n > 2/ε - 1. So for the three values of ε given we can take N = 19, 199, 1999.
   For the second sequence, |(2n²+1)/(n²+ 1) - 2| = 1/(n²+ 1) and for this to be < ε we must take n > √(1/ε - 1) which gives N = 3, 9, 99.

5. (a) If (a_n) = (1, -1, 1, -1, ... ) then (|a_n|) converges even though (a_n) does not.
   (b) If (a_n)→ α then (|a_n|)→|α| as is esy to check from the definition.
   (c) Take a_n= -b_n= n. Then (a_n), (b_n) do not converge, but (a_n+ b_n) does.
   (d) Since (a_n+ b_n) and (a_n) converge, so does the sequence (a_n+ b_n- a_n) = (b_n).

6. If (a_n) converges, the kth decimal place α must be close to (and hence equal to) the (k+1)st, (k+2)nd, etc. That is, the decimal expansion must end in either repeating 0's, repeating 1's, ... , repeating 9's. Such an α is an integer multiple of ¹/₉± something with a terminating decimal expansion.
   The limit of the sequence will then be of the form ^k/₉ with k an integer in the range 0, ... , 9. (The cases k = 0, 9 are 0.0 and 1.0 respectively.)