If x = 1 + 1/2^2 + 1/3^2 + ... + 1/100^2, which of the following must

If \(x = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{100^2}\), which of the following must be true?

A. 0 < x < 1
B. 1 < x < 2
C. 2 < x < 3
D. 3 < x < 4
E. 4 < x < 5

In this x > 1 always
The fraction part after that would always be less than 1 because the highest value of the fraction is \(\frac{1}{2^2}\)(0.25) - almost sum to 1 but not 1.

Hence, 1 + (~1) < 2
Therefore,
1 < x < 2

Answer B.

X is a sum of positive fractions, starting with 1 and adding smaller fractions for each term
Also less than 2, because the second term s
1/2^2=1/4 ,which is less than 1 Continuing this pattern, x is less than 3, 4, and 5.

IMO,
B. 1 < x < 2

OA is B

Is there another approach to this question?­ 
Are there properties for the sum of reciprocals with even denominator?

­Using logic and approximation is the best way to solve this problem.

- Since:
x = 1 + 1/2^2 + 1/3^2 + ... + 1/100^2
x < 1 + 1/(1*2) + 1/(2*3) + ... + 1/(99*100)
x < 1 + (1 - 1/2) + (1/2 - 1/3) + ... + (1/99 - 1/100) => we cancel out 1/2 .... 1/99
x < 2 - 1/100
x < 2­

- ­x > 1, obvious but similar to approach above, we can conclude that x > 3/2
x > 1 + 1/2 - 1/101
x > 3/2

=> Answer is 1 < x < 2­

­An average GMAT aspirant (who is not from math background) would mostly know AP and GP to solve Sequences and Series. Assuming that, we can try to correlate given Series with what we know. Like a Series in Geometric Progression 1 + 1/2 + 1/4 + 1/8 + 1/16 ... Where First Term = 1 and Common Ratio = 1/2

Sum of infinite terms of above GP = (First Term)/(1- Common Ratio) = 1 / (1 - 1/2) = 2
If we compare terms of the above series with the terms of given series the given series would be smaller than this GP we can take as a reference (up to 6 terms and then the values will be too small to matter). Hence, we can conclude that the given series is less that 2 and as something positive is added to 1 it has to be more than 1.

Hence, it is 1 < x <2

Or the other approach would be as given in the picture 
 ­

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