PureMath.pdf

# 4 if φ n and ψ n oscillates infinitely then φ n ψ

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4. If φ ( n ) + and ψ ( n ) oscillates infinitely, then φ ( n )+ ψ ( n ) may tend to + or oscillate infinitely, but cannot tend to a limit, or to -∞ , or oscillate finitely. For ψ ( n ) = { φ ( n ) + ψ ( n ) } - φ ( n ); and, if φ ( n ) + ψ ( n ) behaved in any of the three last ways, it would follow, from the previous results, that ψ ( n ) → -∞ , which is not the case. As examples of the two cases which are possible, consider (i) φ ( n ) = n 2 , ψ ( n ) = ( - 1) n n , (ii) φ ( n ) = n , ψ ( n ) = ( - 1) n n 2 . Here again the signs of + and -∞ may be permuted throughout. 5. If φ ( n ) and ψ ( n ) both oscillate finitely, then φ ( n )+ ψ ( n ) must tend to a limit or oscillate finitely. As examples take (i) φ ( n ) = ( - 1) n , ψ ( n ) = ( - 1) n +1 , (ii) φ ( n ) = ψ ( n ) = ( - 1) n .

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[IV : 65] LIMITS OF FUNCTIONS OF A 152 6. If φ ( n ) oscillates finitely, and ψ ( n ) infinitely, then φ ( n ) + ψ ( n ) oscillates infinitely. For φ ( n ) is in absolute value always less than a certain constant, say K . On the other hand ψ ( n ), since it oscillates infinitely, must assume values numerically greater than any assignable number ( e.g. 10 K , 100 K , . . . ). Hence φ ( n ) + ψ ( n ) must assume values numerically greater than any assignable number ( e.g. 9 K , 99 K , . . . ). Hence φ ( n ) + ψ ( n ) must either tend to + or -∞ or oscillate infinitely. But if it tended to + then ψ ( n ) = { φ ( n ) + ψ ( n ) } - φ ( n ) would also tend to + , in virtue of the preceding results. Thus φ ( n ) + ψ ( n ) cannot tend to + , nor, for similar reasons, to -∞ : hence it oscillates infinitely. 7. If both φ ( n ) and ψ ( n ) oscillate infinitely, then φ ( n )+ ψ ( n ) may tend to a limit, or to + , or to -∞ , or oscillate either finitely or infinitely. Suppose, for instance, that φ ( n ) = ( - 1) n n , while ψ ( n ) is in turn each of the functions ( - 1) n +1 n , { 1 + ( - 1) n +1 } n , -{ 1 + ( - 1) n } n , ( - 1) n +1 ( n + 1), ( - 1) n n . We thus obtain examples of all five possibilities. The results 1–7 cover all the cases which are really distinct. Before passing on to consider the product of two functions, we may point out that the result of Theorem I may be immediately extended to the sum of three or more functions which tend to limits as n → ∞ . 65. B. The behaviour of the product of two functions whose behaviour is known. We can now prove a similar set of theorems con- cerning the product of two functions. The principal result is the following. Theorem II. If lim φ ( n ) = a and lim ψ ( n ) = b , then lim φ ( n ) ψ ( n ) = ab. Let φ ( n ) = a + φ 1 ( n ) , ψ ( n ) = b + ψ 1 ( n ) ,
[IV : 65] POSITIVE INTEGRAL VARIABLE 153 so that lim φ 1 ( n ) = 0 and lim ψ 1 ( n ) = 0. Then φ ( n ) ψ ( n ) = ab + 1 ( n ) + 1 ( n ) + φ 1 ( n ) ψ 1 ( n ) . Hence the numerical value of the difference φ ( n ) ψ ( n ) - ab is certainly not greater than the sum of the numerical values of 1 ( n ), 1 ( n ), φ 1 ( n ) ψ 1 ( n ). From this it follows that lim { φ ( n ) ψ ( n ) - ab } = 0 , which proves the theorem. The following is a strictly formal proof. We have | φ ( n ) ψ ( n ) - ab | 5 | 1 ( n ) | + | 1 ( n ) | + | φ 1 ( n ) || ψ 1 ( n ) | . Assuming that neither a nor b is zero, we may choose n 0 so that | φ 1 ( n ) | < 1 3 / | b | , | ψ 1 ( n ) | < 1 3 / | a | , when n = n 0 . Then | φ ( n ) ψ ( n ) - ab | < 1 3 + 1 3 + { 1 9 2 / ( | a || b | ) } , which is certainly less than if < 1 3 | a || b | . That is to say we can choose n 0 so that | φ ( n ) ψ ( n ) - ab | < when n = n 0 , and so the theorem follows. The reader should supply a proof for the case in which at least one of a and b is zero.

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