Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc

Hint 12.5Show that |S2n− Sn| > 1 2.Hint 12.6 The alternating harmonic series is conditionally convergent.. ln n !n Since the terms in the series on the right side do not vanish as n → ∞,

X

n=0

1ln(n + 20)Shift the indices

∞

X

n=0

(Logπ2)nThis is a geometric series

3 −1

4 =

1121

5 −1

6 =

130

· · ·

Hint 12.5

Show that

|S2n− Sn| > 1

2.Hint 12.6

The alternating harmonic series is conditionally convergent Let {an} and {bn} be the positive and negative terms inthe sum, respectively, ordered in decreasing magnitude Note that both P∞

2 Integrate the series for 1/z.

3 Differentiate the series for 1/z

4 Integrate the series for Log z

Hint 12.21

Evaluate the derivatives of ez at z = 0 Use Taylor’s Theorem.Write the cosine and sine in terms of the exponential function.Hint 12.22

cos z = − cos(z − π)sin z = − sin(z − π)Hint 12.23

∀ > 0 ∃N s.t n > N ⇒ |an| < This is exactly the Cauchy convergence criterion for the sequence {an} Thus we see that limn→∞an = 0 is a necessarycondition for the convergence of the series P∞

n=0an.Solution 12.2

Z ∞ 2

The sum converges.

Z ∞ 10

1

x ln x ln(ln x)dx =

Z ∞ ln(10)

1

y ln y dy =

Z ∞ ln(ln(10))

1

zdzSince the integral diverges, the series also diverges

X

n=0

1ln(n + 20) =

− ln

12n + 1

3n+ 4n+ 5

5n− 4n− 3

1/n

= lim

n→∞

45

(3/4)n+ 1 + 5/4n

1 − (4/5)n− 3/5n

= lim

n→∞

(n + 1)2

((n + 1)2− n2)!

= lim

n→∞

... 1)!)2< /small>(n2< /small>)!((n + 1)2< /small>)!(n!)2< /small>

= lim

n→∞

(n + 1)2< /small>

((n + 1)2< /small>−... data-page="15">

< 12< /p>Trang 16< /span>

Solution 12. 5

Since

|S2n− Sn|... 1)2< /small>

((n + 1)2< /small>− n2< /small>)!

= lim

n→∞

(n + 1)2< /sup>(2n + 1)!

= 0The series is convergent