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The Real And Complex Number Systems

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So, L 1 5 since L ≥ 1.<br />

2<br />

Remark: (1) <strong>The</strong> sequence b n is the famous sequence named Fabonacci sequence.<br />

<strong>The</strong>re are many researches around it. Also, it is related with so called Golden Section,<br />

5 −1<br />

0.618....<br />

2<br />

(2) <strong>The</strong> reader can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s by G.<br />

H. Hardy and E. M. Wright, Chapter X. <strong>The</strong>n it is clear by continued fractions.<br />

(3) <strong>The</strong>re is another proof. We write it as a reference.<br />

Proof: (STUDY) Sinceb n2 b n b n1 , we may think<br />

x n2 x n x n1 ,<br />

and thus consider x 2 x 1. Say and are roots of x 2 x 1, with . <strong>The</strong>nlet<br />

F n n − n<br />

− ,<br />

we have<br />

F n b n .<br />

So,itiseasytoshowthatL 1 5 . We omit the details.<br />

2<br />

Note: <strong>The</strong> reader should be noted that there are many methods to find the formula of<br />

Fabonacci sequence like F n . For example, using the concept of Eigenvalues if we can<br />

find a suitable matrix.<br />

Series<br />

8.15 Test for convergence (p and q denote fixed rela numbers).<br />

(a) ∑ <br />

n1<br />

n 3 e −n<br />

Proof: ByRoot Test, wehave<br />

So, the series converges.<br />

(b) ∑ <br />

n2<br />

log n p<br />

lim n→<br />

sup<br />

n3 1/n<br />

e 1/e 1.<br />

n<br />

Proof: We consider 2 cases: (i) p ≥ 0, and (ii) p 0.<br />

For case (i), the series diverges since log n p does not converge to zero.<br />

For case (ii), the series diverges by Cauchy Condensation <strong>The</strong>orem (or Integral<br />

Test.)<br />

(c) ∑ n1<br />

p n n p (p 0)<br />

Proof: ByRoot Test, wehave<br />

lim n→<br />

sup<br />

pn 1/n<br />

n p.<br />

p<br />

So, as p 1, the series diverges, and as p 1, the series converges. For p 1, it is clear<br />

that the series ∑ n diverges. Hence,<br />

and<br />

<br />

∑<br />

n1<br />

p n n p converges if p ∈ 0, 1

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