Prove that for any n,
\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...... + \frac{1}{n^2}<1
Note:This is not a very strict inequality since the exact value of this limit as n goes to infinity is 0.645
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1 Answers
1This is trivial ...
consider the terms in the expansion from
1/(2k)2 .... 1/(2k+1-1)2
now each term in the series is less than equal to the first term ...
let Sk be their sum
Sk<=1/(2k)2 + 1/(2k)2+ ..... a total of 2k+1-2k terms ..
so
Sk <=(2k+1-2k)/(2k)2 <= 1/2k (cancelling 2k from numerator and denominator)
now Sn = S1 + S2 + S3+ .... Sk + .. some remainder term ....
≤S1+S2+.....+S(k+1)
≤1/2 + 1/4 + 1/8 + .... + 1/2k+1
≤1/2 + 1/4 ... upto infinity
≤1/2/(1-1/2)
≤1