# Math Series Part VIII: N-th Powers of ½

Updated on November 30, 2016

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## An Infinite Couple: Summing to Two

We know, through a little manipulation, that ½ can be rewritten as 2-1. It is with this result in mind that we are then able to calculate the nth power of ½ by setting f(n) = 2-n. Hence, for f(0), we get 1; for f(1), we get ½. For f(2), we get ¼. This allows us to notice that as n tends to infinity, f(n) tends to zero because the denominator is getting larger and larger. The picture to the left is a fitting example of how, because we are working with multiples of two, the next term in the sequence is half of the preceding term—meaning that f(n + 1) = ½ f(n). This tells us that the limit of f(n) as n tends to infinity is 0, which also means that the sum of f(n), starting from n=1, as n tends to infinity, is 1. With this information, we are able to see that since f(0) = 1, the sum of 2-n, n ≥ 0, is 2.

Similarly, if we were not to manipulate ½ at all and simply try to find the nth powers of f(x) = ½x, we would see again that as x gets larger and larger, starting from x=0, we would have the sum of 1/1 + ½ + ¼ + 1/8 + 1/16 + …. and so on, with the denominator growing exponentially as n gets larger. Since we know that the limit approaches zero and that the sum of the terms for n>0 is one, the sum here is also 2.

We can try to prove this more generally by first assuming that there is a sum for this series of expansion of (1/2)n, and we can call this S. We have proven above that S = 1 + 1/2 + 1/4 + 1/8 + 1/16 + … We now have created an equation in S, which we will call E1, that is infinitely long. However, if we take this equation and multiply it by 1/2, we end up with an equation called E2, which is 0.5S = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + … Because the RHS is infinitely long and tends to zero, dividing it by two does little to change the value.

Now, if we perform the subtraction of E1 – E2, we will get a very interesting result that helps us define the value of S in a much more definite way: The subtraction effectively cancels out the 1/2, 1/4, 1/8, 1/16… all the way to the very last value that tends towards zero. This leaves us with the result of S – 0.5S = 1. And this is simplified to show us the some of the series expansion of (1/2)n: S = 2.

This technique of finding the infinite sum through subtraction of two similar equations can be used to find the sum of any geometric series, especially of ones in the form of (1/2)n, and it has demonstrated to us here the simplest way to find the infinite sum of (1/2)n.

If we concentrate now instead on finding a precise formula for calculating Sn = ½ + ½2 + ½3 + … + ½n for any integer n ≥ 1, we will be able to get a simplified result for Sn, which is the sum of the successive first nth powers of ½. Let us achieve this first algebraically. We know that 1 + x + x2 + x3 + … + xn = [1 - xn+1]/(1 - x). We can also prove this by multiplying each side by (1 – x), which gives us (1- x)( 1 + x + x2 + x3 + … + xn) = . Expanding LHS, we get (1 + x + x2 + x3 + … + xn) – (x + x2 + x3 + … + xn + xn+1). Cancelling, we get 1 - xn+1 = RHS.

If we now let x = ½, we get ½ + ½2 + ½3 + … + ½n = 2 – (½)n. We have hereby found a simplified expression for Sn: Sn = 2 – (½)n, and have effectively computed the sum of the successive first nth powers of ½.

We are also able to prove this result with an induction proof. If we consider the LHS of 1 + ½ + ½2 + ½3 + … + ½n = 2 – (½)n, for the smallest value of n (n = 1), we will see that the left-hand side has only two terms: 1 + ½ = 3/2. The RHS when n = 1 is: 2 – ½ = 3/2. Therefore, we have proved that our equation works for n = 1. If we test for n = 2, we get LHS = 1 + ½ + ¼ = 7/4. RHS = 2 – ½2 = 7/4. This implies LHS = RHS and that the test works for n = 2.

Our goal is to now try to test this equation for any value of n, such that n = k. If we suppose that the statement is true for n = k, we get 1+ ½ + ½2 + ½3 + … + ½k = 2 – ½k. Now our task is to prove that this statement is true for n = k + 1, to show that for any subsequent values the statement also holds true. By doing this, we get: 1+ ½ + ½2 + ½3 + … + ½k + ½k+1 = [1+ ½ + ½2 + ½3 + … + ½k] + ½k+1 = , from above, [2 – ½k]+ ½k+1 = 2 - ½k + ½(½k) = 2 – [(½k)(1 – ½)] = 2 - ½(½k) = 2 - ½k+1. This is equal to the RHS for n = k + 1, and so we have proved by induction that ½ + ½2 + ½3 + … + ½n = 2 – ½n for n ≥ 1.

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