The sum of 1/4,1/9,1/16....1/n2 is
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(1)
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less than 1
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(2)
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greater than 1
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(3)
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equal to 1 or less than 1
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(4)
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equal to 1/n2 (n + 2)
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The right answer is Option (1).
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¼ + 1/9 + 1/16 +…+ 1/n2
Consider ¼: ¼ = 1/(2 x 2) < 1/(1 x 2) because ½ < 1 Similarly, 1/9 = 1/(3 x 3) < 1/(2 x 3) Since 1/3 < ½ 1/16 = 1/(4 x 4) < 1/(3 x 4) Because ¼ < 1/3 …………………………………………………….. …………………………………………………….. 1/n2 = 1/(nxn) < 1/(n – 1)n Since 1/n < 1/(n – 1) ¼ + 1/9 + …+ 1/n2 < 1/(1 x 2) + 1/(2 x 3) + … + 1/[(n – 1)n] i.e. ¼ + 1/9 + 1/16 + … + 1/n2 < (1/1 – ½) + (1/2 – 1/3) +…+ [1/(n – 1) – 1/n] i.e. ¼ + 1/9 + 1/16 +… + 1/n2 < 1/1 – 1/n which is less than 1. Hence, the given sum is less than 1 only. Answer: (1)
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Monday, September 24, 2012
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