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February 23

Math puzzle

TOday I did a math competition and on it there was posed a question: WHat is the least possible integer that can be the sum of an infinite geometric sequence {a_n} whose first term is 10. The answer was 6, which I did not get. I got as far as that the sum is $\sum _{i=0}^{\infty }10k^{i}$ (where k is the constant ratio between a_i and a_i-1). After factoring 10 out I grappled with various approaches (even including some Taylor theory and Leibniz' theorem) but I could not figure out what it would be. How did they get 6? Thanks. 72.128.95.0 (talk) 00:24, 23 February 2011 (UTC)[reply]

The geometric series a + ar + ar^2 + ar^3 + ... has the total a/(1-r) when |r| < 1. So in this case, a = 10, and we're trying to minimise the 1/(1-r) term subject to a/(1-r) being an integer. If r = -1, then the "sum" is 10/(1 - -1) = 10/2 = 5, but the series itself doesn't properly converge (in standard arithmetic). However, if r = -2/3, then the series converges and the sum is 6. A bit of graphing or basic calculus can be used to show that this is the minimal solution in the range -1 < r < 1. Confusing Manifestation(Say hi!) 00:37, 23 February 2011 (UTC)[reply]

To add to Confusing Manifestation's solution (in particular, no graphing or calculus needed), note that you have |r| < 1, and for some integer n, 10 = n(1 - r) by rearranging the sum of a geometric series. |r| < 1 implies 0 < 1 - r < 2. Substitute 10/n for (1 - r) and you're left with finding the smallest integer n such that 0 < 1/n < 1/5. Invrnc (talk) 04:29, 23 February 2011 (UTC)[reply]