The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

1) k > 10
2) k < 19

WRITE IT OUT UNTIL YOU SEE A PATTERN.

S(1) = 1 - 1/2
S(2) = 1/2 - 1/3
S(3) = 1/3 - 1/4
etc.

Sum of the first 3 terms:
S(1) + S(2) + S(3) = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) = 1 - 1/4 = 3/4.

When the terms are added, every value except the first and the last CANCELS out.
The result is that, when n=3, the sum = n/(n+1) = 3/4.
Thus, when n=k, we can DEDUCE that the sum = k/(k+1).

Question rephrased: Is k/(k+1) > 9/10?

Statement 1: k>10.
If k=11, then the sum = k/(k+1) = 11/12, which is GREATER than 9/10.
If k=12, then the sum = k/(k+1) = 12/13, which is GREATER than both 9/10 and the preceding sum, 11/12.

As k increases, so does the sum.
Since the LEAST possible sum here = 11/12, the sum will always be GREATER than 9/10.
SUFFICIENT.

Statement 2: k<19.
As shown in statement 1, if k=11, then the sum is GREATER than 9/10.
If k=1, then the sum = k/(k+1) = 1/2, which is LESS than 9/10.

Since the sum in the first case is GREATER than 9/10 and the sum in the second case is LESS than 9/10, INSUFFICIENT.

The correct answer is A.