Why do we have that $ \sqrt{ 1 + \frac{- \log \log n}{\log n} + o(\tfrac{1}{\log n})} = 1 + \frac{- \log \log n}{2\log n} + o(\tfrac{1}{\log n}) $?

Question

I don't understand, why the following is true: $$ \sqrt{ 1 + \frac{- \log \log n}{\log n} + o\left(\tfrac{1}{\log n}\right)} = 1 + \frac{- \log \log n}{2\log n} + o\left(\tfrac{1}{\log n}\right). $$ I know that the $1/2$-factor on the RHS comes from the series expansion of the Binomial series. But I don't see why the little $o$-factor remains the same. Shouldn't the new $o$-factor on the RHS be $$ o\left(\tfrac{\log \log n}{\log n}\right) + o\left(\tfrac{1}{\log n}\right) $$ ?

This problem is taken from the book "Extremes and related properties of random sequences and processes" form Leadbetter, Lindgren, Rootzén. It seems like it is an error.

Answer 1

The correct one is \begin{align} \sqrt{ 1 + \frac{-\log \log n}{\log n} + o\left(\frac{1}{\log n}\right)} &= 1 + \frac{1}{2}\left(\frac{-\log \log n}{\log n} + o\left(\frac{1}{\log n}\right)\right)+o\left(\frac{-\log \log n}{\log n} + o\left(\frac{1}{\log n}\right)\right)\\ &= 1 + \frac{1}{2}\left(\frac{-\log \log n}{\log n} + o\left(\frac{1}{\log n}\right)\right)+o\left(\frac{\log \log n}{\log n} \right)\\ &= 1 - \frac{\log \log n}{2\log n} + o\left(\frac{1}{\log n}\right)+o\left(\frac{\log \log n}{\log n} \right)\\ &= 1 - \frac{\log \log n}{2\log n} +o\left(\frac{\log \log n}{\log n} \right).\\ \end{align}

Answer 2

Note that by definition

$$o(\tfrac{\log \log n}{\log n}) + o(\tfrac{1}{\log n})=o(\tfrac{\log \log n}{\log n}) $$

indeed, with $\omega_i(n)\to 0$

$$o(\tfrac{1}{\log n})=\omega_1(n)\cdot\tfrac{1}{\log n}$$

$$o(\tfrac{\log \log n}{\log n})=\omega_2(n)\cdot\tfrac{\log \log n}{\log n}$$

then

$$o(\tfrac{\log \log n}{\log n}) + o(\tfrac{1}{\log n})=\omega_1(n)\cdot\tfrac{1}{\log n}+\omega_2(n)\cdot\tfrac{\log \log n}{\log n}=\omega_3(n)\tfrac{\log \log n}{\log n}$$

indeed

$$\omega_3(n)=\omega_1(n)\cdot\tfrac{1}{\log \log n}+\omega_2(n)\to 0$$

Note also that we can write correctly for the little-o term

$$o(\tfrac{\log \log n}{\log n}) + o(\tfrac{1}{\log n})$$

while the given little-o term

$$o(\tfrac{1}{\log n})$$

is uncorrect indeed let

$$o(\tfrac{\log \log n}{\log n}) + o(\tfrac{1}{\log n})=\omega_1(n)\cdot\tfrac{1}{\log n}+\omega_2(n)\cdot\tfrac{\log \log n}{\log n}=\omega_4(n)\tfrac{1}{\log n}$$

then

$$\omega_4(n)=\omega_1(n)+\omega_2(n)\cdot \log \log n$$

but we can't conlcude that

$$\omega_2(n)\cdot \log \log n \to 0$$