Evaluating Indefinite Sums

Date: 12/07/2003 at 14:06:25
From: Neil 
Subject: Evaluating indefinite sums

Should I use the comparison test or ratio test or a combination of
these to evaluate the sum

       Inf         
     Sum      1/(3n+1)(4n+2)  
      n=-Inf  

Date: 12/15/2003 at 10:56:42
From: Doctor Pete
Subject: Re: Evaluating indefinite sums

Hi Neil,

The comparsion and ratio tests for series are tests for convergence,
not methods of summation, and are useless to find the exact value of
this sum, although they can easily demonstrate convergence.

That said, the actual summation of this series is rather tricky and 
difficult, and the only way I know how to do it involves some rather 
sophisticated manipulation.  While it is correct, the steps outlined 
below do not constitute a complete proof.

I will be using notation consistent with Mathematica.  Thus,

     Integrate[F[x], {x,a,b}]

will mean the definite integral of the function F[x] from x=a to x=b.
 Similarly,

     Sum[F[x], {x,a,b}]

will denote the sum of F[x] from x=a to x=b.

We are interested in evaluating the related sum (after removing the 
factor of 1/2)

     2S = Sum[1/((2k+1)(3k+1)), {k,-Infinity,Infinity}].

The first step is to introduce a parameter t, and recognize that

     Integrate[t^(3k), {t,0,1}] = 1/(3k+1),  k > -1/3.

Then the partial sum

     S[+] = Sum[1/((2k+1)(3k+1)), {k,0,Infinity}]
          = Sum[Integrate[t^(3k), {t,0,1}]/(2k+1), {k,0,Infinity}],

and since the sum is absolutely convergent, we may switch the order of
summation to produce

     S[+] = Integrate[Sum[t^(3k)/(2k+1), {k,0,Infinity}], {t,0,1}].

Now recall that the Taylor series for ArcTanh[x] about x = 0 gives

     ArcTanh[x] = Sum[x^(2k+1)/(2k+1), {k,0,Infinity}].

(This can be derived in a number of ways, one of which uses the
familiar arctangent series.)  Hence

     S[+] = Integrate[Sum[t^(3/2)^(2k+1)/(2k+1),
                {k,0,Infinity}]/t^(3/2), {t,0,1}]

          = Integrate[ArcTanh[t^(3/2)]/t^(3/2), {t,0,1}].

Thus we have converted part of S into an hyperbolic arctangent
integral on [0,1].

Similarly, we may observe that

     Integrate[t^(-3k-2), {t,0,1}] = -1/(3k+1),  k < -1/3.

Note the sign of the right-hand side term is negative.  So the 
corresponding "negative part" of the sum S is

   S[-] = -Integrate[Sum[t^(-3k-2)/(2k+1), {k,-Infinity,0}, {t,0,1}],

and again applying the hyperbolic arctangent formula, we find

     S[-] = Integrate[ArcTanh[t^(3/2)]/t^(1/2), {t,0,1}].

Thus,

     2S = S[+] + S[-]

        = Integrate[(1+t)ArcTanh[t^(3/2)]/t^(3/2), {t,0,1}].

We have now successfully converted the given sum into a single
definite integral, and now we must compute its antiderivative, which 
is very tedious.  Integrate by parts, with the choice

     u  = ArcTanh[t^(3/2)],

     du = (3/2)t^(1/2)/(1-t^3) dt,

     dv = (1+t)/t^(3/2) dt,

     v  = 2(t^(1/2) - t^(-1/2)).

This gives the product

     uv = 2(Sqrt[t] - 1/Sqrt[t])ArcTanh[t^(3/2)],

and we evaluate the limit at the endpoints, t = 0, t = 1:

     uv[1] = 0,
     uv[0] = 0.

Thus,

     2S = -3 Integrate[Sqrt[t](Sqrt[t]-1/Sqrt[t])/(1-t^3), {t,0,1}]

        = -3 Integrate[(t-1)/(1-t^3), {t,0,1}]

        = 3 Integrate[1/(t^2 + t + 1), {t,0,1}]

        = 3 Integrate[1/((t + 1/2)^2 + (Sqrt[3]/2)^2), {t,0,1}],

which is a simple arctangent integral, whose antiderivative is

        2 ArcTan[(2t+1)/Sqrt[3]]/Sqrt[3].

Evaluating at the endpoints t = 0, t = 1, we find

     2S = 3(2 Pi/(3 Sqrt[3]) - Pi/(3 Sqrt[3]))

        = Pi/Sqrt[3],

from which we finally obtain the desired sum,

     S = Pi/(2 Sqrt[3]).

Some commentary on this solution.  First, not all computations are 
written out in full detail; if you would like to do this as an
exercise, I highly encourage it.  

Second, while it is tedious, it is actually elementary, using no ideas
beyond basic calculus.  

Finally, while I have given you the steps, I haven't really explained
how those steps were chosen; for instance, why not instead use the 
parameter t to rewrite 1/(2k+1) rather than 1/(3k+1)?  The answer is 
because doing so would result in a hypergeometric sum which is not 
easily evaluated further down the line.  In fact, the usage of a 
parameter is a common technique for evaluating complicated sums and 
integrals.

If you have any further questions about this problem, feel free to 
ask!

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