Question by Summer Clouds - Discuss with MathYouLike

Summer Clouds moderators

06/07/2017 at 14:11

Give \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) and \(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\).
Compare A and B.

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Kayasari Ryuunosuke Coordinator 06/07/2017 at 15:33

We have :

\(\dfrac{1}{1^2}>\dfrac{1}{1.2}\)

\(\dfrac{1}{2^2}>\dfrac{1}{2.3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)

..............

\(\dfrac{1}{100^2}>\dfrac{1}{99.100}\)

=> \(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}>\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{99.100}=A\)

So B > A
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