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by GMATGuruNY » Sun Jul 08, 2018 2:03 am
BTGmoderatorDC wrote:$$If\ S\ =\ 1\ +\ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2},\ which\ of\ the\ following\ is\ true?$$

A. S > 3
B. S = 3
C. 2 < S < 3
D. S = 2
E. S < 2
BALLPARK.
After the first four terms, the values are so small that they can be ignored.
Approximate the sum of the first four terms:
$$If\ S\ ≈\ 1\ +\ \frac{1}{4}+\frac{1}{10}+\frac{1}{15} = \ 1\ +\ \frac{15}{60}+\frac{6}{60}+\frac{4}{60} = \ 1\ +\ \frac{25}{60}$$

The resulting sum indicates that S < 2.

The correct answer is E.
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by Brent@GMATPrepNow » Sun Jul 08, 2018 4:14 am
BTGmoderatorDC wrote:$$If\ S\ =\ 1\ +\ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2},\ which\ of\ the\ following\ is\ true?$$

A. S > 3
B. S = 3
C. 2 < S < 3
D. S = 2
E. S < 2
So, S = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + . . . .

Now, let's convert a few of the fractions to decimal approximations...
S = 1 + 0.25 + 0.11 + 0.06 + 0.04 + . . .

Add the first 4 values....
S = 1.42 + 0.04 + . . .

IMPORTANT: Notice that each fraction is less than the fraction before it
So, each of the 5 decimals after 0.04 will be less than 0.04
So, if we replace all of those 5 decimals with 0.04, our new sum will be greater than the original sum

So: S < 1.42 + 0.04 + 0.04 + 0.04 + 0.04 + 0.04 + 0.04

Simplify: S < 1.42 + 0.24
Simplify: S < 1.66

Answer: E

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