Пример 28. Последовательность (xn) определяется следующим образом:
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(n = 2, 3, . . .). |
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Доказать существование предела последовательности (xn) и найти его.
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Решение. Покажем, что n N : 0 < xn < 1. Доказывать это утверждение будем методом математической индукции.
I. При n = 1 неравенство 0 < x1 = 12 < 1 верно. II. Пусть имеет место неравенство
0< xn−1 < 1.
III.Тогда, очевидно, что
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Итак, последовательность (xn) ограниченная.
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Покажем, далее, что последовательность (xn) возрастающая. Действительно
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xn |
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> 0 |
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для всех n = 1, 2, 3, . . . . Следовательно, в силу теоремы 23, последовательность (xn) сходится. Обозначим lim xn = a.
Тогда, переходя к пределу в равенстве
1 x2 − xn = 2 + n2 1,
получим a = 12 + a22. Последнее уравнение имеет два решения: a1,2 = 1. Ответ: lim xn = 1.
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Пример |
29. Доказать, что последователь- |
ность |
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x0 |
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2 + x0 |
2 + x1 |
2 + xn−1 |
где x0 > 0 – произвольное число, сходится и найти предел этой последовательности.
Пример 30. Дана последовательность
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. . . , xn = u2 + |
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n радикалов{z |
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Доказать, что эта последовательность имеет предел, и найти его.
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2.2.10. Число e.
Рассмотрим последовательность с общим чле-
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1 n |
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ном xn = 1 + n |
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и покажем, что она имеет конечный предел. В силу теоремы 23, для доказательства суще-
ствования конечного предела этой последовательности достаточно показать что:
1.последовательность (xn) возрастающая;
2.последовательность (xn) ограничена сверху.
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Пользуясь формулой бинома Ньютона, запишем:
xn = |
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1 · 2 |
1 · 2 · 3 |
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Пользуясь формулой бинома Ньютона, запишем:
xn = |
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1 · 2 |
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n(n − 1)(n − 2) · · · [n − (n − 1)] |
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•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
1. Покажем, что последовательность (xn) -
возрастающая. Для этого запишем выражение для xn+1 и сравним его с xn :
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xn = |
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− n + 1 |
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•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
xn = |
1 + |
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(n + 1)! |
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· · · |
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− n + 1 |
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