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Finding the Sum of Arithmetico-Geometric Series

Date: 09/13/2004 at 13:21:30
From: Sudheer
Subject: Sum of inifinite series

Find the sum of the infinite series

  1/7 + 4/(7^2) + 9/(7^3) + 16/(7^4) +...

I would also like to know if there is a general rule to find the sum
of (n^2/p^n) for n = 1 to infinity.

The series looks like a convergent sequence.  I wonder if there is a 
general rule to break a convergent sequence like above into one or
more convergent sequences?



Date: 09/13/2004 at 13:43:33
From: Doctor Vogler
Subject: Re: Sum of inifinite series

Hi Sudheer,

Thanks for writing to Dr. Math.  Have you ever summed an 
"arithmetico-geometric series"?  That is, one of the form

       k
  S = sum a * n * r^n  ?
      n=1

Some people do this by taking derivatives, but I won't assume that you
know calculus, and I'll do it the algebraic way:  Subtract r*S.

             k                 k
  S - r*S = sum a * n * r^n - sum a * n * r^(n+1)
            n=1               n=1

             k                k+1
  S - r*S = sum a * n * r^n - sum a * (n-1) * r^n
            n=1               n=2

                            k+1
  S - r*S = a * 1 * r^1 + [ sum a * r^n ] - a * k * r^(k+1)
                            n=2

and then the middle series is a geometric series, and I assume that
you can sum a series like that.  This gives you a formula for S - r*S,
and then you just divide both sides by 1-r to get S.

Well, you have a series of the form

       k
  S = sum a * n^2 * r^n
      n=1

In fact, you can sum it in exactly the same way:  Subtract r*S, 
regroup the sums, divide by 1-r, and you will get a formula for S 
which has an arithmetico-geometric series on the right, the type that
I already solved.  Use the formula for that kind of series to get the
formula for your kind of series.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 09/16/2004 at 12:34:39
From: Sudheer
Subject: Thank you (Sum of inifinite series)

Thanks very much.  Your explanation has indeed made things much
simpler for me.  I got this problem in one of the practice tests (for
admission into Indian Institute of Management).  I could solve the
problem using your approach now, however it took more than 2 minutes,
which is more than the maximum time I can allocate for a difficult
question like this in the actual test.  I would like to know if there 
exists a faster method to attack such problems.



Date: 09/16/2004 at 14:58:54
From: Doctor Vogler
Subject: Re: Thank you (Sum of inifinite series)

Hi Sudheer,

You could always memorize the formula for

  inf
  sum i^2 * r^i.
  i=1

Some people like memorizing things.  Others don't.  Alternately, I
mentioned calculus.  You can derive the formulas from

     1      inf
  ------- = sum x^i   (eq. 1)
   1 - x    i=0

by taking the derivative of both sides, which gives

       1        inf               inf
  ----------- = sum i * x^(i-1) = sum (i+1) * x^i
   (1 - x)^2    i=1               i=0,

and then subtracting eq. 1 from this gives

       x        inf
  ----------- = sum i * x^i  (eq. 2)
   (1 - x)^2    i=0

and taking another derivative gives

     1 + x      inf                 inf
  ----------- = sum i^2 * x^(i-1) = sum (i+1)^2 * x^i
   (1 - x)^3    i=1                 i=0

and then you can subtract eq. 1 and twice eq. 2 from this to get

    x^2 + x     inf
  ----------- = sum i^2 * x^i  (eq. 3)
   (1 - x)^2    i=0

and so on....  Or some people just take derivatives and leave their
sums in the form

     (n-1)!     inf
  ----------- = sum (i+1)(i+2)(i+3)...(i+n-1) * x^i
   (1 - x)^n    i=0

and then take linear combinations of those to get a sum of a 
polynomial in i times x^i.

In each case, the sum converges when

  |x| < 1

and diverges when

  |x| > 1.

I'll let you think about the convergence at x = 1, x = -1, and (if you
do complex numbers) at other complex values with absolute value 1.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 09/18/2004 at 03:07:31
From: Sudheer
Subject: Sum of inifinite series

Thank you for the detailed explanation.  I now observe that it is the 
degree of the numerator that actually makes the solving process 
complicated.  Taking the best from algebra and calculus, I've formed 
my own strategy to find the sum of a convergent series like n^2/r^n.  

   a.  Reduce the given sequence into a simple APGP using the 
algebriac method and find the value of  S(1-r).  This would leave us 
with an intermediate series on the R.H.S.

   b.  In this step I simply apply calculus to find the sum of the 
infinite series to find the sums of intermediatary series obtained as 
a result of step a.

  When r < 1
  
  lim  a + (a + d)r + (a + 2d) r^2 + ... = a/(1-r) + rd/(1-r)^2  
 n->oo
            
Most of the questions I get on convergent series in my paper either 
have a degree 1 or 2, i.e of the form n/r^n where I can solve the 
question using Step (b) alone or n2/r^n where I need to use both step
(a) and (b). 

Thanks once again.
Associated Topics:
College Calculus
High School Calculus
High School Sequences, Series

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