Question

sum of infinite of G.P 1/3+1/6+1/12+1/24+......is....

Answer 1

If we call the sum S, then

S = 1/3 + 1/6 + 1/12 + 1/24 + …, and, multiplying by 1/2

(1/2) S = 1/6 + 1/12 + 1/24 + … . Subtracting the 2nd equation from the 1st, we get

S - (1/2) S = 1/3 (the other terms all cancel, thank you), or (1/2) S = 1/3, so S = 2/3.

Of course, there is a well-known formula for this result, namely

S = a/(1-r), where a is the first term in the series and r is the common ratio. (The absolute value of r must be < 1 and unequal to 0). The general result is derived exactly as we found our particular result. In this case, a = 1/3 and r = 1/2, so

S = (1/3)/[1-(1/2)] = (1/3)/(1/2) = 2(1/3) = 2/3.