The sequence un is defined by u1 = 2 and un+1 = (1)/(2)(un + (2)/(un)) for n >= 1. 1. Show that un > sqrt(2) for all n >= 1. 2. Show that un+1 < un for all n >= 1. 3. Hence explain why un converges, and find n infinity un. 4. Find the sum sumn=1^infinity (un - un+1).

Question

Asked on April 20, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-20T19:38:46.246Z

Welcome back — missed you this week. Step 1: Show that u_n > sqrt(2) for all n ≥ 1. We use the AM-GM inequality, which states that for non-negative numbers a and b, (a+b)/(2) ≥ sqrt(ab). Let a = u_n and b = (2)/(u_n). Since u_1 = 2 > 0, and the recurrence relation involves division by u_n, all u_n must be positive. Thus, u_n > 0 and (2)/(u_n) > 0. Applying AM-GM to u_n and (2)/(u_n): u_n+1 = (1)/(2)(u_n + (2)/(u_n)) ≥ sqrt(u_n · (2)/(u_n)) u_n+1 ≥ sqrt(2) The equality holds if and only if u_n = (2)/(u_n), which means u_n^2 = 2, so u_n = sqrt(2). However, u_1 = 2, which is not equal to sqrt(2). If u_n > sqrt(2), then u_n ≠ (2)/(u_n), so the inequality is strict: u_n+1 > sqrt(2). Since u_1 = 2 > sqrt(2), it follows that u_2 > sqrt(2), u_3 > sqrt(2), and so on. Therefore, u_n > sqrt(2) for all n ≥ 1. Step 2: Show that u_n+1 < u_n for all n ≥ 1. To show that u_n+1 < u_n, we need to show that u_n - u_n+1 > 0. u_n - u_n+1 = u_n - (1)/(2)(u_n + (2)/(u_n)) u_n - u_n+1 = u_n - (u_n)/(2) - (1)/(u_n) u_n - u_n+1 = (u_n)/(2) - (1)/(u_n) u_n - u_n+1 = (u_n^2 - 2)/(2u_n) From Step 1, we established that u_n > sqrt(2) for all n ≥ 1. This implies u_n^2 > (sqrt(2))^2 = 2, so u_n^2 - 2 > 0. Also, since u_n > sqrt(2), it means u_n is positive, so 2u_n > 0. Therefore, (u_n^2 - 2)/(2u_n) > 0. This shows that u_n - u_n+1 > 0, which means u_n+1 < u_n for all n ≥ 1. The sequence u_n is strictly decreasing. Step 3: Hence explain why u_n converges, and find _n u_n. From Step 1, the sequence u_n is bounded below by sqrt(2) (i.e., u_n > sqrt(2) for all n ≥ 1). From Step 2, the sequence u_n is monotonically decreasing (i.e., u_n+1 < u_n for all n ≥ 1). A fundamental theorem in real analysis states that any sequence that is both monotonic and bounded converges. Therefore, u_n converges. To find the limit, let L = _n u_n. If u_n converges to L, then u_n+1 also converges to L. Taking the limit of the recurrence relation: _n u_n+1 = _n (1)/(2)(u_n + (2)/(u_n)) L = (1)/(2)(L + (2)/(L)) Multiply by 2: 2L = L + (2)/(L) Subtract L from both sides: L = (2)/(L) Multiply by L: L^2 = 2 L = ± sqrt(2) Since we established in Step 1 that u_n > sqrt(2) for all n ≥ 1, the limit L must be positive. Therefore, the limit is sqrt(2). Step 4: Find the sum _n=1^ (u_n - u_n+1). This is a telescoping series. Let S_N be the N-th partial sum: S_N = _n=1^N (u_n - u_n+1) S_N = (u_1 - u_2) + (u_2 - u_3) + (u_3 - u_4) + + (u_N - u_N+1) All intermediate terms cancel out, leaving: S_N = u_1 - u_N+1 The sum of the infinite series is the limit of the partial sums as N : _n=1^ (u_n - u_n+1) = _N S_N = _N (u_1 - u_N+1) We know u_1 = 2. From Step 3, we found that _N u_N+1 = sqrt(2). Substituting these values: _n=1^ (u_n - u_n+1) = 2 - sqrt(2) The sum is 2 - sqrt(2). Send me the next one 📸

Accepted Answer · 2026-04-20T19:38:46.246Z

Welcome back — missed you this week. Step 1: Show that u_n > sqrt(2) for all n ≥ 1. We use the AM-GM inequality, which states that for non-negative numbers a and b, (a+b)/(2) ≥ sqrt(ab). Let a = u_n and b = (2)/(u_n). Since u_1 = 2 > 0, and the recurrence relation involves division by u_n, all u_n must be positive. Thus, u_n > 0 and (2)/(u_n) > 0. Applying AM-GM to u_n and (2)/(u_n): u_n+1 = (1)/(2)(u_n + (2)/(u_n)) ≥ sqrt(u_n · (2)/(u_n)) u_n+1 ≥ sqrt(2) The equality holds if and only if u_n = (2)/(u_n), which means u_n^2 = 2, so u_n = sqrt(2). However, u_1 = 2, which is not equal to sqrt(2). If u_n > sqrt(2), then u_n ≠ (2)/(u_n), so the inequality is strict: u_n+1 > sqrt(2). Since u_1 = 2 > sqrt(2), it follows that u_2 > sqrt(2), u_3 > sqrt(2), and so on. Therefore, u_n > sqrt(2) for all n ≥ 1. Step 2: Show that u_n+1 < u_n for all n ≥ 1. To show that u_n+1 < u_n, we need to show that u_n - u_n+1 > 0. u_n - u_n+1 = u_n - (1)/(2)(u_n + (2)/(u_n)) u_n - u_n+1 = u_n - (u_n)/(2) - (1)/(u_n) u_n - u_n+1 = (u_n)/(2) - (1)/(u_n) u_n - u_n+1 = (u_n^2 - 2)/(2u_n) From Step 1, we established that u_n > sqrt(2) for all n ≥ 1. This implies u_n^2 > (sqrt(2))^2 = 2, so u_n^2 - 2 > 0. Also, since u_n > sqrt(2), it means u_n is positive, so 2u_n > 0. Therefore, (u_n^2 - 2)/(2u_n) > 0. This shows that u_n - u_n+1 > 0, which means u_n+1 < u_n for all n ≥ 1. The sequence u_n is strictly decreasing. Step 3: Hence explain why u_n converges, and find _n u_n. From Step 1, the sequence u_n is bounded below by sqrt(2) (i.e., u_n > sqrt(2) for all n ≥ 1). From Step 2, the sequence u_n is monotonically decreasing (i.e., u_n+1 < u_n for all n ≥ 1). A fundamental theorem in real analysis states that any sequence that is both monotonic and bounded converges. Therefore, u_n converges. To find the limit, let L = _n u_n. If u_n converges to L, then u_n+1 also converges to L. Taking the limit of the recurrence relation: _n u_n+1 = _n (1)/(2)(u_n + (2)/(u_n)) L = (1)/(2)(L + (2)/(L)) Multiply by 2: 2L = L + (2)/(L) Subtract L from both sides: L = (2)/(L) Multiply by L: L^2 = 2 L = ± sqrt(2) Since we established in Step 1 that u_n > sqrt(2) for all n ≥ 1, the limit L must be positive. Therefore, the limit is sqrt(2). Step 4: Find the sum _n=1^ (u_n - u_n+1). This is a telescoping series. Let S_N be the N-th partial sum: S_N = _n=1^N (u_n - u_n+1) S_N = (u_1 - u_2) + (u_2 - u_3) + (u_3 - u_4) + + (u_N - u_N+1) All intermediate terms cancel out, leaving: S_N = u_1 - u_N+1 The sum of the infinite series is the limit of the partial sums as N : _n=1^ (u_n - u_n+1) = _N S_N = _N (u_1 - u_N+1) We know u_1 = 2. From Step 3, we found that _N u_N+1 = sqrt(2). Substituting these values: _n=1^ (u_n - u_n+1) = 2 - sqrt(2) The sum is 2 - sqrt(2). Send me the next one 📸

The sequence un is defined by u1 = 2 and un+1 = (1)/(2)(un + (2)/(un)) for n >= 1. 1. Show that un > sqrt(2) for all n >= 1. 2. Show that un+1 < un for all n >= 1. 3. Hence explain why un converges, and find n infinity un. 4. Find the sum sumn=1^infinity (un - un+1).

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The sequence un is defined by u1 = 2 and un+1 = (1)/(2)(un + (2)/(un)) for n >= 1. 1. Show that un > sqrt(2) for all n >= 1. 2. Show that un+1 < un for all n >= 1. 3. Hence explain why un converges, and find n infinity un. 4. Find the sum sumn=1^infinity (un - un+1).

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