Let $a_n$ be a sequence of real numbers that only has 2 partial limits: $2$ and $-1$. Define $b_n = \frac{2a_n^2-a_n-1}{a_n^2+1}$.

Prove $b_n$ converges.

It is obvious to me that if we plug in the partial limits of $a_n$ into $b_n$ we get $1$, but how do we prove $b_n$ doesn't actually have more partial limits other than $1$?


1 Answer 1


Hint: $$ |b_n-1|=\bigg|\frac{a_n^2-a_n-2}{a_n^2+1}\bigg|=\bigg|\frac{(a_n-2)(a_n+1)}{a_n^2+1}\bigg|\le|(a_n-2)(a_n+1)| $$ and $\{a_n\}$ is bounded.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .