Mathematical Methods for Physics and Engineering - Matematica.NET

SERIES AND LIMITS◮Sum the seriesN∑(n +1)(n +3).n=1The nth term in this series isu n =(n +1)(n +3)=n 2 +4n +3,and therefore we can writeN∑N∑(n +1)(n +3)= (n 2 +4n +3)n=1n=1N∑N∑ N∑= n 2 +4 n + 3n=1n=1n=1= 1 N(N + 1)(2N +1)+4× 1 N(N +1)+3N6 2= 1 6 N(2N2 +15N + 31). ◭4.2.6 Transformation of seriesA complicated series may sometimes be summed by transforming it into afamiliar series for which we already know the sum, perhaps a geometric seriesor the Maclaurin expansion of a simple function (see subsection 4.6.3). Varioustechniques are useful, and deciding which one to use in any given case is a matterof experience. We now discuss a few of the more common methods.The differentiation or integration of a series is often useful in transforming anapparently intractable series into a more familiar one. If we wish to differentiateor integrate a series that already depends on some variable then we may do soin a straightforward manner.◮Sum the seriesS(x) =x43(0!) + x54(1!) + x65(2!) + ··· .Dividing both sides by x we obtainS(x)x = x33(0!) + x44(1!) + x55(2!) + ··· ,which is easily differentiated to give[ ]d S(x)= x2dx x 0! + x31! + x42! + x53! + ··· .Recalling the Maclaurin expansion of exp x given in subsection 4.6.3, we recognise thatthe RHS is equal to x 2 exp x. Having done so, we can now integrate both sides to obtain∫S(x)/x = x 2 exp xdx.122