Solution 5

1. (a) (2n/(n²+ 1)) =((²/_n)/(1 + ¹/_n²))→ 0/(1 + 0) = 0.
   (b) (n² - 2n + 1)/(n² + 2n + 1) = (1 - 2/n + 1/n²)/(1 + 2/n + 1/n²) and so (a_n)→ (1 - 0 + 0)/(1 + 0 + 0) = 1.
   (c) 3ⁿ/(2ⁿ + 3ⁿ) = 1/((²/₃)ⁿ + 1) and so (a_n)→ 1/(0 + 1) = 1
   (d) √(n + 1) - √n = [(√(n + 1) - √n)(√(n + 1) + √n)]/(√(n + 1) + √n) = [(n + 1) - n]/(√(n + 1) + √n) = 1/(√(n + 1) + √n) and so the sequence→ 0 as n→ ∞.

2. (a) Taking n = 1, 3, 7, 9, 13, 15, ... gives a subsequence (^√3/₂ , 0 , ^√3/₂ , 0 , ... ) which does not converge. Hence the original sequence does not converge.
   (b) The subsequence of even terms→ 1 and the subsequence of odd terms→ -1 and so (since the limits of subsequences of a convergent sequence are the same as the limit of the sequence) the sequence is not convergent.

3. 0 < n!/nⁿ = (1.2.3. ... .n)/(n.n. ... .n) < ¹/_n. Since the constant sequence (0)→ 0 and the sequence (¹/_n)→ 0 the given sequence→ 0 by the "squeeze rule".

   Stirling's formula shows that n! grows approximately like √(2π) n^n+0.5e^-n. For example, for n = 100, n! is about 0.93326 × 10¹⁵⁸ while Stirling's approximation is 0.93248 × 10¹⁵⁸. An even better approximation is obtained by replacing e^-n by e^-n+1/(12n). This gives an approximation of 0.93323 × 10¹⁵⁸ for 100!.

4. a_n+1- a_n= 2a_n/(1+a_n) - 2a_n-1/(1+a_n-1) = 2(a_n- a_n-1)/[(1+a_n)(1+a_n-1)]
   Hence, provided a_n-1, a_n> -1, the sequence is monotonic.
   If a₁ = 2 then a₂ = ⁴/₃ and so the sequence is monotonic decreasing and since it is clearly bounded below by 0, it is convergent.
   If a₁ = ¹/₂ then a₂ = ²/₃ and so the sequence is monotonic increasing. It is easy to prove by induction that if a₁ < 1 then a_n < 1 for all n and so the sequence is bounded above and hence convergent.
   In either case, the limit α must satisfy α = 2α/(1 + α) ⇒ α = 1.

5. Given ε > 0, choose N so that a_2n is within ε of α for 2n > N and a_2n-1 is within ε of α for 2n - 1 > N. Then all the terms of the sequence are eventually within ε of α and the sequence converges.

6. The sequence (r_n) is (1, 2, 1.5, 1.6, 1.625, 1.615, 1.619, 1.618, ... ) (approximately). Hence it is not monotonic -- though you might notice that the subsequences of odd and even terms are monotonic.
   r_n+1 = f_n+2/f_n+1 = (f_n+1 + f_n)/f_n+1 = 1 + f_n/f_n+1= 1 + 1/r_n.
   So if the sequence (r_n) converges to a limit α, α satisfies α = 1 + 1/α ⇒ α²- α - 1 = 0 ⇒ (if (α > 1) α = (√5 + 1)/2.
   From the above, r_n+2 = 1 + 1/r_n+1 = 1 + 1/(1+1/r_n) = 1 + r_n/(r_n+ 1) = (2r_n+ 1)/(r_n+ 1) as required.
   r_n+2 - r_n= (2r_n + 1)/(r_n + 1) - (2r_n-2 + 1)/(r_n-2 + 1) = 2(r_n - r_n-2)/[(r_n + 1)(r_n-2 + 1)].
   Hence by induction, since r₃ - r₁ > 0, the sequence of odd terms is monotonic increasing and since r₄ - r₂ < 0, the subsequence of even terms is monotonic decreasing. Hence, by question 4 above, the sequence converges to (√5 + 1)/2.

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