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tgt_2k9

Which is larger:

2 / 201 or ln(101/100)

gaurav_cheetah

2/101

edison

ln(101/100) > 2 / 201

tgt_2k9

plzzzzzz solve completly

hpudipeddi

2/201<101/100
by cross multiplying
2*100<>201*101
Since 201*101 is greater
201*101>2*100
i.e.101/100>2/201

saidvait

ln101/100<2/201

hsbhatt

$\frac{2}{201} = \frac{1}{100 + \frac{1}{2}} = \frac{1}{100 \left(1 + \frac{1}{200} \right)} \\ \\ = \frac{1}{100} \left(1 - \frac{1}{200} + \frac{1}{200^2} - \frac{1}{200^3} + ...\infty \right ) \\ \\ = \frac{1}{100} \left (1 - \frac{1}{2 \times 10^2} + \frac{1}{4 \times 10^4} - \frac{1}{8 \times 10^6} +...\infty \right)$

$\ln \frac{101}{100} = \ln \left( 1 + \frac{1}{100} \right) \\ \\ = \frac{1}{100} - \frac{1}{2 \times 100^2} + \frac{1}{3 \times 100^3} - \frac{1}{4 \times 100^4} + ... \infty \\ \\ \text{(Using the Taylor expansion for} \ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... \infty ) \\ \ = \frac{1}{100} \left ( 1 - \frac{1}{2 \times 10^2} + \frac{1}{3 \times 10^4} - \frac{1}{4 \times 10^6} + ... \infty \right)$

Comparing the expansions, you can see that they differ from the 3rd term onwards. The 3rd term of the ln expansion is greater and the terms following are of much lower order and hence can be ignored.

Thus we conclude that $\ln \frac{101}{100} > \frac{2}{201}$

feynmann

." The 3rd term of the ln expansion is greater and the terms following are of much lower order and hence can be ignored."

Not conclusive !!

hsbhatt

Of course it is conclusive.

The ln series is greater than the fraction at the 3rd term in the order of 10^-6.

Subsequently, the difference is decreased but in the order of 10^-8. Any further decrease occurs in the order 10^-12 and so on.

So, if you write it out the negative contribution in decimal terms, you are adding digits at the 8th place and then in the 12h place and so on, you never reach order 8.

This is not counting the positive contributions in alternate terms.

That is why, in one of the representations of the Taylor series, you write do not write beyond x⁴ term. It is just abbreviated to O(x⁴) (See wikipedia).

It just takes a bit of logical thinking to see how this is true.

hsbhatt

And I will be happy if someone comes up with a more conclusive proof or can prove convincingly that my proof is not conclusive

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