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The first two terms are 1 and 1/2. So the sum should be definitely greater than 1½ . For the sum to be greater than or equal to 2, the sum of next fractions should be at least 1/2. Now the succeeding fractions are less than 1/2, hence irrespective of the fraction, the sum will be less than 1/2. So, sum of the sequence will be greater than 1½ but not more than 2.

For all positive integers n, the nth term in sequence S_n is defined [#permalink]

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10 Sep 2015, 10:31

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This post received KUDOS

Bunuel wrote:

For all positive integers n, the nth term in sequence \(S_n\) is defined as follows:

\(S_n = (n!)^{-1}\)

The sum of the first six terms of \(S_n\) is

(A) between 0 and ½ (B) between ½ and 1 (C) between 1 and 1½ (D) between 1½ and 2 (E) greater than 2

Kudos for a correct solution.

from the sequence equation we can see first 6 terms to be \(1+ \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} +\frac{1}{720}\) So from 1st two numbers we have sum 1.5 and adding other four will be give more than 1.5 so A, B , C gone. From D and E we can easily select D as from last four terms we have \(\frac{1}{6} (1+ \frac{1}{4} + \frac{1}{20} +\frac{1}{120})\) which is less than .5. Hence D
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The sum of just the first two terms, by the way, is 1½. Glancing ahead at the answers, we should already be able to rule out A, B, and C at this point. The terms never go negative, so the only question is whether the sum of the first six terms takes us above 2 or not.

The sum of the first six terms can be written this way:

Given how quickly the fractions shrink, we might guess at this point that the sum does not go above 2. However, we can make sure of this guess. Leaving out the first two terms (the 1 and the ½), we can just figure out the sum of the last four terms:

Re: For all positive integers n, the nth term in sequence S_n is defined [#permalink]

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21 Aug 2017, 12:01

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