The probability of choosing an integer ‘$k$’ out of $2m$ integers $1,2,3,…2m$ is inversely proportional to $k^4$.If $P_1$ is the probability that chosen number is odd and $P_2$ is the probability that chosen number is even, then
A. $P_1 < \frac{2}{3}$
B. $P_1 > \frac{1}{2}$
C. $P_1 > \frac{2}{3}$
D. $P_2 < \frac{1}{2}$
It is obvious that B and D are true, since probability of choosing odd number is greater than choosing even number. After trying on for some values of $m$, I feel C is also true, although the official answer is BD. The value of $P_1$ will be $\sum_{r=1} ^ m \frac{x}{(2r-1)^4}$ where $x= \frac{1}{\sum_{k=1}^{2m} k^{-4}}$.
I tried to prove C by taking $\displaystyle{\lim_{m \to \infty}} P_1$ but couldn't compute it. Wolfram Alpha says that it's equal to $\frac{15}{16}$. Is there some other way to prove option C?
For $m=1$ you have $P_1 = \frac{\frac1{1^4}}{\frac1{1^4}+\frac1{2^4}} = \frac{16}{17}> \frac23$
Now consider the position as $m \to \infty$.
Let $S_\infty= \sum\limits_{k=1}^\infty \frac{1}{k^4}$ so $\frac{S_\infty}{2^4} = \sum\limits_{k=1}^\infty \frac{1}{(2k)^4}$ and $P_1 \to \frac{S_\infty-\frac{S_\infty}{2^4}}{S_\infty} = \frac{15}{16}> \frac23$ as $n$ increases
This suggests that $C$ is true. For a proof, you can adjust the previous part:
Let $S_m= \sum\limits_{k=1}^{2m} \frac{1}{k^4}$ so $\frac{S_m}{2^4} = \sum\limits_{k=1}^{2m} \frac{1}{(2k)^4} > \sum\limits_{k=1}^{m} \frac{1}{(2k)^4}$ so for all $m$ we have $$P_1 = \frac{\sum\limits_{k=1}^{2m} \frac{1}{k^4}- \sum\limits_{k=1}^{m} \frac{1}{(2k)^4} }{\sum\limits_{k=1}^{2m} \frac{1}{k^4}} > \frac{S_m-\frac{S_m}{2^4}}{S_m} = \frac{15}{16}> \frac23.$$