IIT JEE , AIEEE Forum - Mathematics , Algebra

Abhiroop

Prove that

1 + (1/2²) + (1/3²) +........+ (1/n²) < 7/4

Plz show how to solve.....

hsbhatt

$1 + \frac{1}{2^2} + \frac{1}{3^2} + ...+\frac{1}{n^2}$

First we can club the terms as

$1 + \left( \frac{1}{2^2} + \frac{1}{3^2} \right)+ \left(\frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2}+ \frac{1}{7^2} \right) + ...+\frac{1}{n^2}$

Next, we can prove that each successive term is less than 1/2 the previous term.

For example,

$\frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2}+ \frac{1}{7^2} < 2 \left(\frac{1}{4^2} + \frac{1}{6^2} \right) \\ \\<br/>=\frac{1}{2} \left(\frac{1}{2^2} + \frac{1}{3^3} \right)$

Now, $\frac{1}{2^2} + \frac{1}{3^2} < \frac{3}{8}$ is easy to prove as 104<108.

Hence the given expression is less than

$1 + \frac{3}{8} + \frac{3}{16} + ....+ \infty \\ \\<br/>= 1 + \frac{3}{8} \left(1+ \frac{1}{2} + \frac{1}{2^2} + ...+ ..\infty} \right) = 1 + \frac{3}{4} \\ \\<br/>= \boxed{\frac{7}{4}}<br/>$

jishnudas1991

I did it This way:

rajatsen91

Well done dipanjan!

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