Find the sum of the infinite series. \(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots \)The answer is not \(3997\over3993\)
+23246 You can work this problem without calculus (but it's very messy!):
1 + (2/x) + (3/x2) + (4/x3) + (5/x4) + ...
= 1 + (1/x + 1/x) + (1/x2 + 2/x2) + (1/x3 + 3/x3) + ...
= [1 + 1/x + 1/x2 + 1/x3 + ... ] + [1/x + 2/x2 + 3/x3 + 4/x4 + ...]
= [1 + 1/x + 1/x2 + 1/x3 + ... ] + [1/x + (1/x2 + 1/x2) + (1/x3 + 2/x3)+ (1/x4 + 3/x4) + ...]
= [1 + 1/x + 1/x2 + 1/x3 + ... ] + [1/x + 1/x2 + 1/x3 + 1/x4 + ...] + [1/x2 + 2/x3 + 3/x4 + ...]
= [Red Part] + [Blue Part] + [1/x2 + (1/x3 + 1/x3) + (1/x4 + 2/x4) + (1/x5 + 3/x5) + ...]
= [Red Part] + [Blue Part] + [1/x2 + 1/x3 + 1/x4 + 1/x5 + ...] + [1/x3 + 2/x4 + 3/x5 + ...]
Etc.
Now, let's see what we have:
[1 + 1/x + 1/x2 + 1/x3 + ... ]
+ [1/x + 1/x2 + 1/x3 + 1/x4 + ...] = (1/x)[1 + 1/x + 1/x2 + 1/x3 + 1/x4 + ...]
+ [1/x2 + 1/x3 + 1/x4 + 1/x5 + ...] = (1/x2)[1 + 1/x + 1/x2 + 1/x3 + ...]
+ ...
=
(1)[1 + 1/x + 1/x2 + 1/x3 + ... ]
+ (1/x)[1 + 1/x + 1/x2 + 1/x3 + 1/x4 + ...]
+ (1/x2)[1 + 1/x + 1/x2 + 1/x3 + ...]
+ (1/x3))[1 + 1/x + 1/x2 + 1/x3 + ...]
+ ...
= [1 + 1/x + 1/x2 + 1/x3 + ...] · [1 + 1/x + 1/x2 + 1/x3 + ...]
= [ 1 / (1 - 1/x) ] · [ 1 / (1 - 1/x) ]
= [ x / (x - 1) ] · [ x / (x - 1) ]
= [ x / (x - 1) ] 2
For this problem: (1998 / 1997 )2
