Sum the first six terms of 1 + 3/4 + 9/16 + 27/64 and so on, as an exact fraction. What share of the infinite sum do those six terms capture?

Sum the first six terms of 1 + 3/4 + 9/16 + 27/64 and so on, as an exact fraction. What share of the infinite sum do those six terms capture?

Approach: Identify the ratio, apply the finite geometric sum formula, and compare the result against the infinite sum to size the remaining tail.

3367/1024. The geometric series has common ratio r = 3/4, so the partial sum of n terms is (1 - r^n)/(1 - r), which here is (1 - (3/4)^6) * 4. Now (3/4)^6 is 729/4096, since 3^6 is 729 and 4^6 is 4096, so the sum is 4 * (4096 - 729)/4096 = 4 * 3367/4096 = 3367/1024, which is 3.2881. The infinite sum is 1/(1 - 3/4) = 4, so six terms capture 82.2% of it and the tail is 4 * 729/4096 = 0.712. Powers of 3 up to 729 and powers of 4 up to 4096 carry most quick series work, and the tail of a geometric sum is always the infinite sum times r^n, which is the fastest way to decide how many terms are enough.

Follow-up: How many terms of the same series are needed before the tail falls below 0.01?

Key concepts: geometric series, partial sum, common ratio.