Evaluating Indefinite SumsDate: 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/ |
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