First we check that this is true for n=1: S1 = 1/(1x3) which is equal to n/(2n+1) for n=1 therefore Sn = n/(2n+1) is true for n = 1. Next assume that it is true for n=k. Sk = k/(2k+1). Now using this assumption we check that it is true for n=k+1: Sk+1 = Sk+ 1/(2(k+1) - 1)(2(k+1)+1). Rearranging this and substituting in k/(2k+1) for Sk we get Sk+1 = (k+1)/(2k+3) which is consistent with the original formula. Therefore we can say that since Sn = n/(2n+1) is true for n=1 and whenever it is true for n=k it is also true for n=k+1, it is true for all integer values of n larger than or equal to 1.
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