Quiz 4 Quiz 4                              Friday, April 27, 2001



NAME:

Student Number:------            No.----           Marks ----






(25 Marks) Determine whether the series k = 1[((k!)2)/(2k)!] is convergent or divergent.

Solution.

Let an = [((k!)2)/(2k)!]. Then an+1 = [(((k+1)!)2)/((2(k+1))!)] and


an+1
an
=
((k+1)!)2
(2(k+1))!
× (2k)!
(k!)2
=
((k+1)!)2
(k!)2
× (2k)!
(2(k+1))!
=
((k+1)k!)2
(k!)2
× (2k)!
(2k+2)(2k+1)(2k)!
=
((k+1))2
(2k+2)(2k+1)
=
(k2+2k+1
(4k2+6k+2)
=
1/4 < 1

It follows from the Ratio Test that k = 1[((k!)2)/(2k)!] is convergent.


File translated from TEX by TTH, version 2.79.
On 12 May 2001, 09:14.