1. Telescoping series:
    4
    (4n-3)(4n+1)
    		 = A
    4n-3
    		 - B
    4n+1
    		 .
    So
    A(4n+1)+B(4n-3) = 4.
    At 3/4 we get 4A = 4 or A = 1 while at n = -1/4 we get -4B = 4 so B = -1. Then the series is

    å
     n 
    		 1
    4n-3
    		 - 1
    4n+1
    or
    ( 1
    1
    		 - 1
    5
    		 )-( 1
    5
    		 = 1
    9
    		 )+¼+( 1
    4n-3
    		 - 1
    4n+1
    		 .
    As n® ¥ we get
    1- 1
    4n+1
    		 ® 1.

  2. 1+3+¼+243 is a geometric series(though also an arithmetic ones so should really have put 9 in after 3). The terms are
    tn = arn-1
    where a = 1 and r = 3. Then 243 = 3n-1 so ln243 = (n-1)ln3 and
    n-1 = ln243
    ln3
    		 = 5.49
    1.098
    		 = 5.
    So n = 6. We could of course have done this by hand and said the sequence is 1,3,9,27,81,243 to get the same result. Then the sum is
    a(rn-1)
    r-1
    		 = 728/2 = 364.
    Again for this we could have added them up to get the same result.


  3. 1.414141414 = 1.4*(1+1/100+1/(100)2+¼
    which is
    1.4*[(1/100)n-1]/[1/100-1] = 1.4* -1
    -99
    100
    		 = 140
    99
    		 .

File translated from TEX by TTH, version 2.70.
On 13 May 2002, 16:00.