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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 \(3997\over3993\)

 May 18, 2020
 #1
avatar+30028 
+2

As follows (using the Feynman technique):

 

 May 18, 2020
 #2
avatar
0

This simplifys to \(3997\over3993\)

Guest May 18, 2020
 #3
avatar+30028 
+2

No! It doesn't:

 

\(\frac{3992004}{3988009}=1.001001753005071...\\...\\ \frac{3997}{3993}=1.0010017530678688...\)

 

Close, but no cigar!

Alan  May 18, 2020
edited by Alan  May 18, 2020
 #4
avatar+20876 
+1

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

 May 18, 2020

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