Mathematical Methods for Physics and Engineering - Matematica.NET

4.2 SUMMATION OF SERIESIntegrating the RHS by parts we findS(x)/x = x 2 exp x − 2x exp x +2expx + c,where the value of the constant of integration c canbefixedbytherequirementthatS(x)/x =0atx = 0. Thus we find that c = −2 and that the sum is given byS(x) =x 3 exp x − 2x 2 exp x +2x exp x − 2x. ◭Often, however, we require the sum of a series that does not depend on avariable. In this case, in order that we may differentiate or integrate the series,we define a function of some variable x such that the value of this function isequal to the sum of the series for some particular value of x (usually at x =1).◮Sum the seriesS =1+ 2 2 + 3 2 2 + 4 2 3 + ··· .Let us begin by defining the functionf(x)=1+2x +3x 2 +4x 3 + ··· ,so that the sum S = f(1/2). Integrating this function we obtain∫f(x) dx = x + x 2 + x 3 + ··· ,which we recognise as an infinite geometric series with first term a = x and common ratior = x. Therefore, from (4.4), we find that the sum of this series is x/(1 − x). In other words∫f(x) dx =x1 − x ,so that f(x) isgivenbyf(x) = d ( x)1=dx 1 − x (1 − x) . 2The sum of the original series is therefore S = f(1/2) = 4. ◭Aside from differentiation and integration, an appropriate substitution cansometimes transform a series into a more familiar form. In particular, series withterms that contain trigonometric functions can often be summed by the use ofcomplex exponentials.◮Sum the seriesS(θ)=1+cosθ +cos 2θ2!+cos 3θ3!+ ··· .Replacing the cosine terms with a complex exponential, we obtain{}exp 2iθ exp 3iθS(θ) =Re 1+expiθ + + + ···2! 3!}(exp iθ)2 (exp iθ)3=Re{1+expiθ + + + ··· .2! 3!123