In my lecture's note, literally said in the title, there is a question like the below.
$P(x=n) = {4 \over {(n^2 -1)n}}$ for $n =2,3,4,5,...$.
Find the $\lim\limits_{n\to \infty} P(x \leq 2n \vert n<x)$
He doesn't any suggest his solution, only claimed the answer is $3 \over 4$ But, Whenever I did it over and over again, I couldn't got that.
Here is my solution.
$P(x=n) = 2\{ {1\over n-1} + {1\over n+1} - {2\over n}\}$
Then, $P(n \lt x \leq m ) = P(x=n +1) + P(x=n+2) + ... + P(x=m) = 2 \{ {1\over n} - {1\over n+1} - {1\over m} + {1\over m+1}\} $
So, $\lim\limits_{m\to \infty} P(n \lt x \leq m ) = P(n \lt x) = 2 \{ {1\over n} - {1\over n+1}\}$ $= {2 \over {n(n+1)}}$
Plus, $P(n \lt x \leq 2n ) = 2 \{ {1\over n} - {1\over n+1} - {1\over 2n} + {1\over 2n+1}\} = 2 \{ {{n^2+3n+1}\over n(n+1)(2n+1)}\} $
Hence, $\lim\limits_{n\to \infty} P(x \leq 2n \vert n<x) = {{ P(n \lt x \leq 2n ) }\over{P(n \lt x)}} = \infty$
Which point did I have a mistake? Any help would be appreciated.
I carefully speculate his question is non-sense because I can't find any numerical error in my solution. :(
Your calculation cannot be correct as the conditional probability must be bounded above by $1$
Once you have $$P(n \lt x \leq m ) = 2 \left( {1\over n} - {1\over n+1} - {1\over m} + {1\over m+1}\right)$$ I would have thought the next steps could be $$P(n \lt x \leq 2n ) = 2 \left( {1\over n} - {1\over n+1} - {1\over 2n} + {1\over 2n+1}\right)= \frac{3n+1}{n(n+1)(2n+1)}$$ $$P(n \lt x ) = 2 \left( {1\over n} - {1\over n+1} \right)= {2\over n(n+1)} $$ $$P(n \lt x \leq 2n\mid n \lt x ) = \frac{3n+1}{4n+2}$$ and that last expression tends to $\frac34$ as $n$ increases