Solve the logarithmic equation

Question

I have logarithmic problem: $$\frac{\log \:_{10}\left(1\:+\:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\right)}{\log _{10}\left(2+\frac{2}{3}+\frac{2}{9}+...\right)}\cdot \left(\log _2\left(3\right)+\log _3\left(4\right)+\log _{16}\left(3\right)+...+\log _{2^{2n}}\left(3\right)+..\right)$$

I solved first part: $$\frac{\log \:_{10}\left(2\right)}{\log _{10}\left(3\right)}\cdot \left(\log _2\left(3\right)+\log _3\left(4\right)+\log _{16}\left(3\right)+...+\log _{2^{2n}}\left(3\right)+..\right)$$

But can't understand second part.

Answers: $A=2, B=-1, C=-2, D=\frac{1}{2}$

Answer 1

$$ S = \sum_{n=1}^\infty \log_{2^{2n}}(3) = \sum_{n=1}^\infty \log_{4^n}(3) = \sum_{n=1}^\infty \frac{\ln 3}{\ln 4^n} = \frac{\ln 3}{\ln 4} \sum_{n=1}^\infty \frac{1}{n} $$ which diverges.

Answer 2

For the second part we can write for terms after the first two $$\log_{2^{2n}}3=a_n\\ \left(2^{2n}\right)^{a_n}=3\\2^{2na_n}=3\\2^{a_n}=3^{\frac 1{2n}}\\ a_n=\frac 1{2n}\log_23$$ and all the terms after the first two become a diverging harmonic series.

Now that you have added the choices, it is clear the text expects the series to converge. Alpha agrees with me-you can see the screenshot below. I assume the book answer is A. All the terms are positive, so the result cannot be negative. The first two terms of the second part are greater than $1$, so the whole expression is greater than $2\frac {\log_{10}2}{\log_{10}3}\approx 1.26$ which rules out D. Even if it did converge, the irregularity of the first two terms would make me sure the result is not some nice number like $2$.